Definition: Let $\varphi: G \to H$ be a homomoprhism. A fiber of $\varphi$ is a subset of $G$ that maps to a single element in $H$.
Definition: If $\varphi: G \to H$ is a homomorphism, then the kernel of $\varphi$ is the fiber mapping to the identity element in $H$.
Proposition 1: Let $G, H$ be groups and $\varphi: G \to H$ a homomorphism.
Proof: (1) $\varphi(1 \cdot 1) = \varphi(1)\varphi(1) = \varphi(1)$, so $\varphi(1) = 1_H$ by cancellation. $\blacksquare$
(2) $\varphi(1) = \varphi(gg^{-1}) = \varphi(g)\varphi(g^{-1}) = 1$, and multiplying on the left by $\varphi(g)^{-1}$ gives $\varphi(g^{-1}) = \varphi(g)^{-1}$. $\blacksquare$
(3) By induction: suppose $\varphi(g^n) = \varphi(g)^n$. Then $\varphi(g^n \cdot g) = \varphi(g)^n\varphi(g) = \varphi(g^{n + 1})$. And the base case holds for $n = 1$ trivially. $\blacksquare$
(4) $1 \in \operatorname{ker} \varphi$ so it is non-empty. Let $a, b \in \operatorname{ker} \varphi$. Then $\varphi(ab^{-1}) = \varphi(a)\varphi(b^{-1}) = 1$ so $ab^{-1} \in \operatorname{ker} \varphi$. $\blacksquare$
(5) $\varphi(G)$ is non-empty since identity is preserved. Let $a, b \in \varphi(G)$. Then $\varphi(x) = a$ and $\varphi(y) = b$ for some $x. y \in G$. $\varphi(y^{-1}) = b^{-1}$, so $\varphi(xy^{-1}) = ab^{-1}$. $\blacksquare$
Definition: Let $\varphi: G \to H$ be a homomorphism with kernel $K$. The quotient group or factor group, $G/K$ (read $G$ modulo $K$ or $G$ mod $K$) is the group whose elements are the fibers of $\varphi$ with the group operation between two fibers above $a$ and $b$ as the fiber above the product $ab$.
Proposition 2: Let $\varphi: G \to H$ be a homormophism of groups with kernel $K$. Let $X \in G/K$ be the fiber above $a$. Then
Proof: (1) Let $u \in X$. We first show $X \subseteq uK$. Let $g \in X$. Let $k = u^{-1}g$. Then $\varphi(k) = 1$ so $k \in \operatorname{ker} \varphi$, $g = uk$, so $g \in uK$. Thus $X \subseteq uK$. Now let $g \in uK$. Then $\varphi(g) = \varphi(u)\varphi(k) = \varphi(u) = a$ since $u \in X$, so $g \in X$. Thus $uK \subseteq X$. $\blacksquare$
(2) Proceeds symmetrically as (1).
Definition: For any $N \subseteq G$ and any $g \in G$, let
\[gN = \lbrace gn \mid n \in N \rbrace\]and
\[Ng = \lbrace ng \mid n \in N \rbrace\]be the left coset and right coset of $N$ in $G$, respectively. Any element of a coset is called a representative for the coset.
Note that Proposition 2 tells us that $gN = g’N$ for any $g’ \in gN$, motivating the nomenclature of a coset representative.
Lemma 3: Let $N \leq G, u, v \in G$. The following are equivalent:
Proof: $(1 \to 2)$. Suppose $uN = vN$. Then $x = un_u = vn_v$ for some $x \in uN, n_u, n_v \in N$.So $u = vn_vn_u^{-1} \in vN$.
$(2 \to 3)$ Suppose $u \in vN$. Then $u = vn$ for some $n \in N$, so $v^{-1}u = n$, so $v^{-1}u \in N$.
$(3 \to 1)$ Suppose $v^{-1}u \in N$. Let $x \in uN$. Then $x = un$ for some $n \in N$, so $u = xn^{-1}$. Thus $v^{-1}u = vxn^{-1}$, so $x = v^{-1}(v^{-1}u)n$. By the assumption, $v^{-1}u \in N$, so $v^{-1}un \in N$, so $x \in vN$. So $uN \subseteq vN$. Symmetrically, $vN \subseteq uN$, so $uN = vN$.
Theorem 3: Let $G$ be a group and let $K$ be a kernel of some homomorphisms from $G$ to another group. Then the set whose elements are the left cosets of $K$ in $G$ with operation defined by
\[uK \circ vK = (uv)K\]is well-defined.
Proof: First we show the operation is well-defined. Let $uK = u’K$ and $vK = v’K$. By Proposition 2, it suffices to show that $u’v’ \in (uv)K$. By Proposition 2, $u’ \in uK$, so $u’ = uk_u$. Similarily, $v’ = vk_v$. So $u’v’ = (uk_u)(vk_v)$, and $\varphi(u’v’) = \varphi(uk_u)\varphi(vk_v) = \varphi(u)\varphi(v)$. So $u’v’$ and $uv$ are in the same coset of $K$, that is, $u’v’ \in (uv)K$.
Examples:
$(2 = 4)$ $u \in vN$ if and only if $u$ is a representative of $vN$, so $uN = vN$ if and only if $u$ and $v$ are representatives of the same coset. $\blacksquare$
Proposition 4: Let $N \leq G$. The set of left cosets of $N$ form a partition of $G$.
Proof: First we show $\bigcup_{g \in G} gN = G$. Let $x \in gN$. Then $x = gn$ for some $g \in G$, $n \in N \subseteq G$ so $x \in G$ by closure. Now suppose $x \in G$. Since $N \leq G$, $1 \in N$, so $x = x1 \in gN$. This shows subset inclusion in both directions, so $\bigcup_{g \in G} gN = G$.
Next we show that $uN \cap vN = \emptyset$ for $uN \neq vN$. Suppose $uN \cap vN \neq \emptyset$. Then $x = un_u = vn_v$ for some $x \in uN \cap vN$, $n_u, n_v \in N$. In particular, $u = vn_vn_u^{-1} \in vN$, so $uN = vN$. $\blacksquare$
Proposition 5: Let $G$ be a group and $N \leq G$. The operation on left cosets $uN \cdot vN = (uv)N$ is well defined if and only if $gng^{-1} \in N$ for all $g \in G$ and $n \in N$. This operation also defines a group on the set of left cosets of $N$.
Proof: Suppose $gng^{-1} \in N$ for all $g \in G, n \in N$. Then $gn = n’g$ for some $n’ \in N$. We show that if $u’ \in uN$ and $v’ \in vN$,
\[\begin{align*} u'v' &= (un_u)(vn_v) \\ &= uv(n_u'n_v) \\ &\in (uv)N \end{align*}\]whre $n_u, n_v, n’_u \in N$, and the second line uses the assumption.
Now suppose the operation is well-defined. Then for all $u, u’ \in uN$ and $v, v’ \in vN$, $uvN = u’v’N$. Let $g \in G, n \in N$. In particular, choose $u = 1$, $u’ = n$, $v = v’ = g^{-1}$. Note that $u \in u’N$ since $1 \in N$ and $v \in v’N$ since $v = v’$, both by Lemma 3. We have $g^{-1}N = ng^{-1}N$, so $gng^{-1}N = N$, so $gng^{-1} \in N$, by Lemma 3.
That this operation defines a group follows from the associativity of $G$. It can also be shown that the identity is $1N$ and the inverse for some coset $aN$ is $a^{-1}N$. $\blacksquare$
Definition: The element $gng^{-1}$ is called the conjugate of $n \in N$ by $g$. The set $gNg^{-1} = \lbrace gng{-1} \mid n \in N \rbrace$ is called the conjugate of $N$ by $g$. The element $g$ normalizes $N$ if $gNg^{-1} = N$. A subgroup $N \leq G$ is called normal if every element of $G$ normalizes $N$, and is written $N \trianglelefteq G$
Theorem 6: Let $N \trianglelefteq G$. The following are equivalent:
Proof: $(1 \to 2)$ by definition.
$(2 \to 3)$ Suppose $gNg^{-1} = N$ for all $g \in G$. If $gn \in gN$, then $gng^{-1} = n’$ for some $n’ \in N$, so $gn = n’g$ and $gn \in Ng$. So $gN \subseteq Ng$, and symmetrically, $Ng \subseteq gN$. So $gN = Ng$.
$(3 \to 4)$ If $gN = Ng$ then $gng^{-1} \in N$ for all $g \in G, n \in N$, and $(4)$ follows from Proposition 5.
$(4 \to 5)$ If $(4)$, then $gng^{-1} \in N$ by Proposition 5, so $(5)$.
$(5 \to 1)$ Let $n \in N$, $g \in G$. Then $g^{-1}ng = n’$ for some $n’ \in N$, so $n = gn’g^{-1} \in gNg^{-1}$. So $N \subseteq gNg^{-1}$ and $N = gNg^{-1}$, thus $N \trianglelefteq G$, by definition. $\blacksquare$
Proposition 7: A subgroup $N \trianglelefteq G$ if and only if it is the kernel of some homomorphism.
Proof: Suppose $N \trianglelefteq G$. Define $\phi: G \to G/N$ as $\phi(g) = gN$. It is a homomorphism since $\phi(xy) = (xy)N = (xN)(yN) = \phi(x)\phi(y)$, since the operation is well-defined by Proposition 5. Also $\operatorname{ker}(\phi) = N$ since $gN = 1N = N$ if and only if $g \in N$ by Lemma 3.
Now suppose $N$ is the kernel of some homomorphism. We show that $gNg^{-1} \subseteq N$, from which $N \trianglelefteq G$ follows from Theorem 6. Let $gng^{-1} \in gNg^{-1}$. and $\phi$ the homomorphism for which $N$ is a kernel. Then $\phi(gng^{-1}) = \phi(g)\phi(n)\phi(g^{-1}) = \phi(g)\phi(g^{-1}) = 1$, so $gng^{-1} \in \operatorname{ker}\phi = N$.
(For an even shorter proof of this converse, note that by Proposition 2 the left and right cosets of the kernel coincide, thus by $(3)$ in Theorem 6, $N \trianglelefteq G$.) $\blacksquare$
Definition: Let $N \trianglelefteq G$. The homomorphism $\pi: G \to G/N$ defined by $\pi(g) = gN$ is the natural projection (homomorphism) of $G$ onto $G/N$. If $\overline H \leq G/N$ is a subgroup of $G/N$, the complete preimage of $\overline H$ is $\pi^{-1}(\overline H)$.
Examples:
by Exercise 4. Thus, quotient groups of cyclic groups are cyclic.
The exercises below highlight some additional important properties, e.g.
Exercise 1: Let $\phi: G \to H$ be a homomorphism and $E \leq H$. Then $\phi^{-1}(E) \leq G$. If $E \trianglelefteq H$, then $\phi^{-1}(E) \trianglelefteq G$. And, $\operatorname{ker} \phi \trianglelefteq G$.
Proof: $1 \in \phi^{-1}(E)$ since $1 \in E$ and $\phi(1^{-1}) = \phi(1)^{-1}$. So $\phi^{-1}(E)$ is non-empty. Now let $a, b \in \phi^{-1}(E)$. Then $\phi(a), \phi(b) \in E$, so $\phi(a)\phi(b)^{-1} = \phi(ab^{-1}) \in E$, so $ab^{-1} \in \phi^{-1}(E)$.
Now suppose $E \trianglelefteq H$. Let $g \in G$ and $x \in \phi^{-1}(E)$. Then $\phi(gxg^{-1}) \in E$ by normality of $E$ so $gxg^{-1} \in \phi^{-1}(E)$.
Finally, $\operatorname{ker} \phi \trianglelefteq G$ by Proposition 7.
$\blacksquare$
Exercise 2: Let $\phi: G \to H$ be a homomorphism with kernel $K$ and $a, b \in \phi(G)$ with corresponding fibers $X, Y$. Let $u \in X$. Show that if $XY = Z$ in $G/K$ and $w \in Z$, then there is a $v \in Y$ with $uv = w$.
Proof: Since $XY = Z, Y = X^{-1}Z = (u^{-1}K)(wK) = (u^{-1}w)K$. So $u^{-1}w \in Y$. $\blacksquare$
Exercise 3: Let $A$ be an abelian group and $B \leq A$. Then $A/B$ is abelian. Give an example of a non-abelian group containing a proper normal subgroup $N$ with $G/N$ abelian.
Proof: Let $X, Y \in A/B$. Let $x \in X, y \in Y$ Then $XY = xByB = xyB = yxB = yBxB$.
(Note we use Proposition 2 in the claim that $X = xB$ and $Y = yB$). $\blacksquare$
For an example, consider $S_3/\lbrace 1, (12) \rbrace$, which is abelian.
Exercise 4: Prove that in the quotient gruop $G/N, (gN)^\alpha = g^\alpha N$ for all $\alpha \in \mathbb Z$.
Proof: $(gN)(gN) = ggN = g^2N$, and the exercise follows by induction. Negative powers proceed likewise with $g^{-1}$. $\blacksquare$
Exercise 5: Prove the order of $gN$ in $G/N$ is $n$, where $n$ is the smallest positive integer such that $g^n \in N$, and infinite order if no such $n$ exists.
Proof: By Exercise 4, the smallest $n$ such that $(gN)^n = N$ is precisely the smallest such $n$ such that $g^n \in N$. since if $g^n \in N$ then $g^nN = N$ (and likewise for the infinite case).$\blacksquare$
The example holds whenever $g \in N$ but $g \neq 1$. For example, let $G = \mathbb Z_6$ and $N = \mathbb Z_3$. Then the order of $g = 3$ is 2 but the order of $gN = 3 + \mathbb Z_3 = \mathbb Z_3$ is 1.
Exercise 14: Consider the additive quotient group $\mathbb Q/\mathbb Z$.
a. Show that every coset of $\mathbb Z$ in $\mathbb Q$ contains exactly one representative $q \in \mathbb Q$ in the range $0 \leq q \lt 1$.
Proof: Consider $r + \mathbb Z$, $r \in \mathbb Q$. The sole representative is $r - k$ where $k$ is the largest integer smaller than $r$. $0 \leq r - k \lt 1$ since $0 \leq r - k$ since $r$ is larger than $k$ and $r - k \lt 1$ since if it weren’t, then $1 \leq r - k$ and $k + 1$ would be a larger integer smaller than $r$; a contradiction. $\blacksquare$
b. Show that every element of $\mathbb Q/\mathbb Z$ has finite order, but there are elements of arbitrarily large order.
Proof: Every element $a/b + \mathbb Z$ has finite order since $b(a/b) \in \mathbb Z$. The order is $\operatorname{lcm}(a, b)$, so orders can be made arbitrarily large by choosing $a, b$ with arbitrarily large LCMs. $\blacksquare$
c.: Show that $\mathbb Q/\mathbb Z$ is the torsion subgroup of $\mathbb R/\mathbb Z$.
Proof: If $r + \mathbb Z$ where $r \in \mathbb R$ is irrational, then $kr \neq n$ for any $n \in \mathbb Z$ for all $k \in \mathbb Z$ since if it did, then $r = k/n$, contradicting that $r$ is irrational. $\blacksquare$
Exercise 15: Prove that the quotient of a divisible abelian group by any proper subgroup is also divisible.
Proof: Let $G$ be a divisible abelian group with $H \lt G$. Let $X \in G/H$ and $k \in \mathbb Z$. Let $x \in X$ and $y \in Y$ such that $y^k = x$, by divisibility. Then $Y^k = (yH)^k = xH = X$. $\blacksquare$
Exercise 16: Let $G$ be a group and $N \trianglelefteq G$. Define $\overline G = G/N$. Show that if $G = \langle S \rangle$ for any $S \subseteq G$, then $\overline G = \langle \overline S \rangle$.
Proof: Let $X \in \overline G$ and $x \in X$. Then $x$ is some finite product of elements in $S$. So $X$ is some finite product of elements in $\overline S$ since $aNbN = abN$. So $X \in \langle \overline S \rangle$ and $\overline G \subseteq \langle \overline S \rangle$. Similar in the opposite direction. $\blacksquare$
Exercise 36: Prove that if $G/Z(G)$ is cyclic then $G$ is abelian.
Proof: Let $g \in G$. Since $G/Z(G)$ is cyclic, $g \in gZ(G) = x^aZ(G)$ for some $x \in G$ and $a \in \mathbb Z$. So $g = x^az$ for some $z \in Z(G)$. Similarily, $h = x^bz’$. So $gh = x^azx^bz’ = x^ax^bzz’ = x^bz’x^az = hg$. $\blacksquare$
Exercise 42: Let $H, K \trianglelefteq G$ with $H \cap K = 1$. Prove that $xy = yx$ for all $x \in H$ and $y \in K$.
Proof: Let $x \in H, y \in K$. Then $x^{-1}y^{-1}xy = x^{-1}h \in H$ where $h = y^{-1}xy \in H$ by normality of $H$. Similarily, $x^{-1}y^{-1}xy = ky \in K$ where $k = x^{-1}y^{-1}x \in K$ by normality of $K$. So $x^{-1}y^{-1}xy \in H \cap K = 1$, so $xy = yx$. $\blacksquare$
Theorem 8 (Lagrange’s Theorem): If $G$ is a finite group and $H$ is a subgroup of $G$, then the order of $H$ divides the order of $G$, and the number of left cosets of $H$ in $G$ equals $\vert G \vert/\vert H \vert$.
Proof: By Proposition 4, the cosets of $H$ define a partition over $G$, so it suffices to show that $\vert gH \vert = \vert H \vert$ for any coset $gH$. Define $\sigma_g: H \to gH$ as $h \mapsto gh$, which is surjective since $gh = \sigma_g(h)$ and injective since if $gh = gh’$ then $h = h’$ by cancellation. So $\vert G \vert = k\vert H \vert$, where $k$ is the number of cosets of $H$.
Consequently, $k = \vert G \vert / \vert H \vert$. $\blacksquare$
Definition: If $G$ is a group and $H \leq G$, the number of left cosets of $H$ in $G$ is called the index of $H$ in $G$ and denoted $\vert G : H \vert$.
Corollary 9: If $G$ is a finite group and $x \in G$, then the order of $x$ divides the order of $G$.
Proof: $\langle x \rangle$ is a subgroup of $G$, so $\vert \langle x \rangle \vert = \vert x \vert \mid \vert G \vert$. $\blacksquare$
Corollary 10: If $G$ is a group of prime order $p$, then $G$ is cyclic.
Proof: $G$ must not be trivial, so let $x \in G - \lbrace 0 \rbrace$. Then by Lagrange’s, $\vert \langle x \rangle \vert = p$ (not 1 since $x \neq 1$), so $\langle x \rangle = G$.
Example: Any subgroup $H \leq G$ of index 2 is normal in $G$.
Proof: Consider the 2 cosets of $H$, $H$ and $gH$. Since cosets form a partition, $gH = G - H$. Similarily, the 2 right cosets of $H$ are $H$ and $Hg$, and $Hg = G - H$. $gH = Hg$ so $H$ is normal. $\blacksquare$
Theorem 11 (Cauchy’s Theorem): If $G$ is a finite group and $p$ is a prime dividing $\vert G \vert$, then $G$ has an element of order $p$.
Proof: Later. $\blacksquare$
Theorem 12 (Sylow): If $G$ is a finite group of order $p^\alpha m$, where $p$ is prime and $p \nmid m$, then $G$ has a subgroup of order $p^\alpha$.
Proof: Later. $\blacksquare$
Definition: Let $H$ and $K$ be subgroups of a group. Then the direct product $HK = \lbrace hk \mid h \in H, k \in K \rbrace$.
Proposition 13: If $H$ and $K$ are finite subgroups of a group, then
\[\vert HK \vert = \frac{\vert H \vert \vert K \vert}{\vert H \cap K \vert}\]Proof: Note that $HK = \bigcup_{h \in H} hK$. The distinct cosets of $hK$ are the distinct cosets of $h(H \cap K)$, since if $h_1, h_2 \in K$ then $h_1K = h_2K$. By Lagrange’s Theorem, $\vert H : H \cap K\vert = \vert H \vert/\vert H \cap K \vert$. We showed in the proof of Lagrange’s Theorem that $\vert hK \vert = \vert K \vert$. Multiplying the number of distinct cosets by the order of each coset gives the result. $\blacksquare$
Proposition 14: If $H$ and $K$ are subgroups of a group, $HK$ is a subgroup if and only if $HK = KH$.
Proof: Suppose $HK = KH$. First note that $HK$ is non-empty since $H$ and $K$ are non-empty. Let $a, b \in HK$. Then $ab^{-1} = hkk’^{-1}h’^{-1}$ for some $h, h’ \in H$ and $k, k’ \in K$. Since $KH = HK$, $kk’^{-1}h’^{-1} = h^\star k^\star$ for some $h^\star \in H$ and $k^\star \in K$. So $ab^{-1} \in HK$, and $HK$ is a subgroup by the subgroup criterion.
Now suppose $HK$ is a subgroup. Let $x \in HK$. Since $HK$ is a subgroup, $x = (hk)^{-1}$ for some $hk \in HK$. Then $x^{-1} = (h’k’)^{-1} = k’^{-1}h’^{-1} \in KH$, so $HK \subseteq KH$. A symmetric argument shows $KH \subseteq HK$. $\blacksquare$
Definition: If $A \subseteq N_G(K)$ (or $C_G(K)$), then $A$ normalizes $K$ (centralizes $K$).
Corollary 15: If $H$ and $K$ are subgroups of $G$ and $H$ normalizes $K$, then $HK \leq G$.
Proof: Let $hk \in HK$. Since $H$ normalizes $K$, $k = h^{-1}k’h$ for some $k’$, so $hk = hh^{-1}k’h = k’h \in KH$ so $HK \subseteq KH$. Similarily $KH \subseteq HK$, so by Proposition 14, $HK \leq G$. $\blacksquare$
Some properties shown in the exercises:
Exercise 1: Only the orders $n$ that divide 120, and their corresponding index is $120/n$.
Exercise 4: Show that if $\vert G \vert = pq$ for some primes $p, q$ (not necessarily distinct), then $G$ is abelian or $Z(G) = 1$.
Proof: By Lagrange’s theorem, the order of $Z(G)$ is either $pq, p, q$ or 1. If the order is $pq$, then $Z(G) = G$ so $G$ is Abelian. If it is $p$ (or $q$), then $\vert G/Z(G) \vert = q$ (or $p$), so it is cyclic and thus Abelian by Corollary 10. Thus, $G$ is either Abelian or $\vert Z(G) \vert = 1$ so $Z(G) = 1$. $\blacksquare$
Exercise 5: Let $H$ be a subgroup of $G$ and let $g \in G$.
a. Prove that $gHg^{-1}$ is a subgroup of $G$ and of same order as $H$.
Proof: $H’ = gHg^{-1}$ is non-empty since $H$ is non-empty. Let $a, b \in H’$. Then $ab^{-1} = (gxg^{-1})(gyg^{-1}) = gxyg^{-1} \in H’$ for some $x, y \in H$. So $H’$ is a subgroup. To show that $\vert H’ \vert = \vert H \vert$, define the map $\phi: H \to H’$ as $h \mapsto ghg^{-1}$, which can easily be shown to be an isomorphism. $\blacksquare$
b. Deduce that if $n \in \mathbb Z^+$ and $H$ is the unique subgroup of $G$ of order $n$ then $H \trianglelefteq G$.
Proof: By (a), $gHg^{-1}$ is a subgroup of $G$ of order $n$, so $H = gHg^{-1}$. $\blacksquare$
Exercise 6: Let $H \leq G$ and let $g \in G$. Prove that if $Hg$ equals some left coset of $H$, then $Hg = gH$ and $g \in N_G(H)$.
Proof: Suppose $Hg = xH$ where $x \in G$. Note that $g \in gH$ and $g \in Hg$ since $1 \in H$, so $g \in xH$. Since $xH$ and $gH$ share an element, $gH = xH$. So $Hg = gH$ and $g$ normalizes $H$. $\blacksquare$
Exercise 8: Prove that if $H$ and $K$ are finite subgroups of $G$ whose orders are relatively prime then $H \cap K = 1$.
Proof: Since $H \cap K$ is a subgroup of $H$ and of $K$, if $\vert H \cap K \vert = n \neq 1$, then by Lagrange’s, $n$ divides $H$ and $K$, so their orders are not relatively prime. $\blacksquare$
Exercise 10: Suppose $H$ and $K$ are subgroups of finite index $m$, $n$, respectively, in $G$ (possibly infinite). Prove that $\operatorname{lcm}(m,n) \leq \vert G : H \cap K \vert \leq mn$.
Proof: First, note that $a(H \cap K) = aH \cap aK$, so the number of cosets of $H \cap K$ is at most the product of the number of cosets of $aH$ and of $aK$, that is, $mn$. For the lower bound, note that since $H \cap K \leq H$ and $H \cap K \leq K$,
\[\vert G : H \cap K\vert = \vert G : K\vert\vert K : H \cap K\vert = \vert G : H\vert\vert H : H \cap K\vert\]since $H \cap K$ partitions $H$ which partitions $G$ (likewise for $H \cap K$ paritioning $K$ partitioning $G$) (see Exercise 11 for a complete proof). So $m$ and $n$ divide $\vert G : H \cap K\vert$, so $\operatorname{lcm}(m, n) \leq \vert G : H \cap K\vert$. $\blacksquare$
Exercise 11: Let $H \leq K \leq G$. Prove that $\vert G : H \vert = \vert G : K \vert \cdot \vert K : H \vert$.
Proof: Let $n = \vert G : K \vert, m = \vert K : H \vert$. We have
\[\begin{align*} G = \bigsqcup_{g \in G} gK = \bigsqcup_{g \in G}g\bigsqcup_{k \in K}kH = \bigsqcup_{g \in G, k \in K} gkH \end{align*}\]and there are $mn$ distinct cosets of $gkH$. $\blacksquare$
Exercise 12: Let $H \leq G$. Prove that $x \mapsto x^{-1}$ maps each left coset of $H$ in $G$ onto a right coset of $H$ and gives a bijection between the set of left cosets and the set of right cosets of $H$ in $G$.
Proof: Let $g \in G$ and $x \in Hg$. Then $x = hg$ for some $h \in H$, so $x^{-1} = g^{-1}h^{-1} \in g^{-1}H$, so the mapping is surjective. Now suppose $hg = h’g$. Then, taking the inverse of both sides, shows $g^{-1}h^{-1} = g^{-1}h’^{-1}$, so the mapping is injective. Thus it is bijective. $\blacksquare$
Theorem 16 (The First Isomorphism Theorem): If $\phi: G \to H$ is a homomorphism of groups, then $\operatorname{ker} \phi \trianglelefteq G$ and $G/\operatorname{ker}\phi \cong \phi(G)$.
Proof: The first statement is Proposition 7, and the second is Theorem 3 and Proposition 5. $\blacksquare$
Corollary 17: Let $\varphi: G \to H$ be a homomorphism of groups.
Proof: $(1)$ Suppose $\varphi$ is injective. Then for $a \in G$, $\phi(a) = 1$ if and only if $a = 1$ since $\varphi(1) = 1$. Conversely, suppose the kernel is trivial. Then $G/1 = G \cong \varphi(G)$ by the First Isomorphism Theorem, so $\varphi$ is bijective (and thus injective).
$(2)$ $\vert G : \operatorname{ker} \varphi \vert = \vert G/\operatorname{ker} \varphi \vert = \vert \varphi(G) \vert$ since $G/\operatorname{ker} \varphi \cong \varphi(G)$ by the First Isomorphism Theorem. $\blacksquare$.
Theorem 18 (The Second Isomorphism Theorem): Let $G$ be a group, $A, B \leq G$, and $A \leq N_G(B)$. Then $AB \leq G$, $B \trianglelefteq AB$, $A \cap B \trianglelefteq A$, and $AB/B \cong A/A \cap B$.
Proof: $AB \leq G$ since $A \leq N_G(B)$ by Corollary 15.
We have $B \trianglelefteq A$, so $(ab)B(ab)^{-1} = abBb^{-1}a^{-1} \subseteq aBa^{-1}\subseteq B$ for $ab \in AB$, so $B \trianglelefteq AB$ by Theorem 6.
Define $\varphi: A \to AB/B$ as $\varphi(a) = aB$. It is surjective since if $abB \in AB/B$, then $\varphi(a) = abB = aB$. It is also a homomorphism since $\varphi(aa’) = (aa’)B = aBa’B = \varphi(a)\varphi(a’)$. The identity is $1B$, which is mapped to when $a \in B$, so $\operatorname{ker} \varphi = B$. Thus by the First Isomorphism Theorem, $AB/B \cong A/A \cap B$, and $A \cap B \trianglelefteq A$ by Proposition 7. $\blacksquare$
Theorem 19 (The Third Isomorphism Theorem): Let $G$ be a group and $H, K \trianglelefteq G$ with $H \leq K$. Then $K/H \trianglelefteq G/H$ and
\[(G/H)/(K/H) \cong G/K\]Proof: Define a map $\varphi: G/H \to G/K$ as $\varphi(gH) = gK$. It is well-defined since if $gH = g’H$ then $gK = g’K$ since if $g \in H$ then $g \in K$ since $H \leq K$ (and likewise for $g’$). It is surjective since $g$ may be chosen arbitrarily. Finally, $\operatorname{ker} \varphi = K/H$ since if $kH \in K/H$ then $\varphi(kH) = kK = K$ since $k \in K$, and conversely, if $\varphi(gH) = K$ for $gH \in G/H$ then $g \in K$, that is, $gH \in G/K$. So by the First Isomorphism Theorem, the statement holds and $K/H \trianglelefteq G/H$ by Proposition 7. $\blacksquare$
Theorem 20 (Fourth Isomoprhism Theorem) (Correspondence Theorem) (Lattice Isomorphism Theorem): Let $G$ be a group and $N \trianglelefteq G$. Then there is a bijection between the set of subgroups $A$ of $G$ containing $N$ and the set of subgroups $A/N$ of $G/N$. In particular, for all $A, B \leq G$ and $N \leq A, N \leq B$,
where $\overline X := X/N$ for some group $X$.
Proof: Let $X^\star$ be the set of subgroups $A$ of $G$ containing $N$ and $Y^\star$ the set of subgroups $A/N$ of $G/N$. Define $\varphi: X^\star \to Y^\star$ as $\varphi(A) = \pi(A)$. It is surjective since the complete preimage of a subgroup in $G/N$ is a subgroup in $G$ (by Exercise 1 in “Definition and Examples”). It is injective since if $\pi(A) = \pi(A’)$ then $A = \pi^{-1}(\pi(A)) = \pi^{-1}(\pi(A’)) = A’$. So $\varphi$ is bijective.
$(1)$ Suppose $A \leq B$. $\overline A$ is non-empty since $1 \in A$, so $\pi(1) \in \overline A$. Let $aN, a’N \in \overline A$. Then $(aN)(a’N)^{-1} \in \overline A$ since $A$ is a subgroup, so $\overline A \leq \overline B$.
Now suppose $\overline A \leq \overline B$. $A$ is non-empty since $1 \in \overline A$ so $\pi^{-1}(1) \in A$. Let $a, a’ \in A$. Then $aa’^{-1} = \pi^{-1}(aN)\pi^{-1}((a’N)^{-1}) = \pi^{-1}((aa’^{-1})N) \in A$ since $\overline A \leq \overline B$. So $A \leq B$.
$(2)$ Suppose $A \leq B$. Define a mapping from the cosets in $B/A$ to the cosets of $\overline B/\overline A$ as $bA \mapsto (bN)\overline A$. It is injective and well-defined since
\[\begin{align*} bA = b'A &\iff b'^{-1}b \in A \\ &\iff b'^{-1}bN \in \overline A \\ &\iff (bN)\overline A = (b'N)\overline A \end{align*}\]and it is surjective since $B$ ranges over all values of $b$ in $(bN)\overline A$.
$(3)$ We have
\[\begin{align*} \overline X \in \overline{\langle A, B \rangle} &\iff \overline X = (x_1 \dots x_n)N \\ &\iff \overline X = (x_1N)\dots(x_nN) \\ &\iff \overline X \in \langle \overline A, \overline B \rangle \end{align*}\]where $x_1, x_2, \dots, x_n \in A \cup B$.
$(4)$ We have
\[\begin{align*} xN \in \overline{A \cap B} &\iff x \in A \cap B \\ &\iff xN \in \overline A \land xN \in \overline B \\ &\iff xN \in \overline A \cap \overline B. \end{align*}\]$(5)$ Suppose $A \trianglelefteq G$. Let $gN \in \overline G$ and $aN \in \overline A$. Then $(gN)(aN)(gN)^{-1} = (gag^{-1})N = N$. For the converse, suppose $\overline A \trianglelefteq \overline G$ and let $a \in A$ and $g \in G$. Then $gag^{-1} \in \pi^{-1}(gN)\pi^{-1}(aN)\pi^{-1}((gN)^{-1}) = \pi^{-1}(gag^{-1}N) = \pi^{-1}(N) = A$, where $\pi$ is the natural projection.
$\blacksquare$
Proposition: Let $N \trianglelefteq G$ and $\varphi$ a homomorphism on $G/N$ whose value for some coset $gN$ depends only on $g$. Define $\phi$ as a homomorphism on $G$ with $\phi(g) = \varphi(\pi(g))$, where $\pi$ is the natural projection. Then $\varphi$ is well-defined if and only if $N \leq \operatorname{ker}{\phi}$.
Proof: Suppose $N \leq \operatorname{ker}\phi$. Let $gN = g’N \in G/N$. Then $g’^{-1}g \in N$, so $g’^{-1}g \in \operatorname{ker}\phi$, so $\phi(g’^{-1}g) = \varphi(g’^{-1}gN) = 1$, so $\varphi(g’^{-1}N) = \varphi(g^{-1}N)$.
Suppose now that $\varphi$ is well-defined. First note that $N \subseteq \operatorname{ker}\phi$ since if it weren’t, then given $gN = g’N$, one may have $g’^{-1}g \in N - \operatorname{ker} \phi$, so $\phi(g’^{-1}g) =\varphi(g’^{-1}gN) = x \neq 1$, thus $\varphi(g’^{-1}N) = x\varphi(gN)$ with $x \neq 1$, so $\varphi(g’^{-1}N) \neq \varphi(gN)$, violating the well-definedness of $\varphi$. Finally, since $N \trianglelefteq G$, $N$ is a group so $N \leq \operatorname{ker} \phi$.
$\blacksquare$
Exercise 1: Let $F$ be a finite field of order $q$ and let $n \in \mathbb Z^+$. Prove that $\vert GL_n(F) : SL_n(F) \vert = q - 1$.
Proof: Define $\varphi: GL_n(F) \to F - \lbrace 0 \rbrace$ as $\varphi(M) = \det M$. It is a homomorphism since $\det(M)\det(M’) = \det(MM’)$ and it is surjective since given some $d \in F$, a matrix $M \in GL_n(F)$ with identical entries to the identity matrix, except for $M_{00} = d$, has determinant $d$ (the product of the entries along the diagonal). The kernel of $\varphi$ is the set of matrices with determinant 1, that is, $SL_n(F)$. So by the First Isomorphism Theorem, $GL_n(F)/SL_n(F) \cong F$, so $\vert GL_n(F) : SL_n(F) \vert = \vert F - \lbrace 0 \rbrace \vert = q - 1$. $\blacksquare$
Exercise 2: Prove all parts of the Lattice Isomorphism Theorem (proved above).
Exercise 3: Prove that if $H$ is a normal subgroup of $G$ of prime index $p$ then for all $K \leq G$, either ($K \leq H$) or ($G = HK$ and $\vert K : K \cap H \vert = p$).
Proof: Suppose $K \nleq H$. By Corollary 15, $HK \leq G$, so $H \leq HK \leq G$. We have $[G : H] = p = [G : HK][HK : H]$ by Exercise 11 in “More on Cosets and Lagrange’s Theorem”, so either $[G : HK] = 1$, in which case $G = HK$, or $[G : HK] = p$. If $[G : HK] = p$, then $[HK : H] = 1$, but then $HK = H$ and $K \leq HK = H$; a contradiction. So $[G : HK] = 1$ and $G = HK$.
To show that $[K : K \cap H] = p$, note that by the Second Isomorphism Theorem, $HK/H \cong K/(K \cap H)$, and since $[HK : H] = p$, $[K : K \cap H] = p$. $\blacksquare$
Exercise 4: Let $C \trianglelefteq A$ and $D \trianglelefteq B$. Prove that $(C \times D) \trianglelefteq (A \times B)$ and $(A \times B)/(C \times D) \cong (A/C) \times (B/ D)$.
Proof: Define $\varphi: A \times B \to A/C \times B/D$ as $\varphi((a, b)) = (aC, bD)$. It is a surjective homomorphism with kernel $C \times D$, so by the First Isomorphism Theorem and Proposition 7, both statements hold. $\blacksquare$
Exercise 7: Let $M$ and $N$ be normal subgroups of $G$ such that $G = MN$. Prove that $G/(M \cap N) \cong (G/M) \times (G/N)$.
Proof: By the Second Isomorphism Theorem, $G/M = MN/M \cong N/(M \cap N)$ and $G/N = MN/N \cong M/(M\cap N)$. So let $\varphi: M/(M \cap N) \times N/(M \cap N) \to G/(M \cap N)$ be defined as $\varphi: (m(M \cap N), n(M \cap N)) \mapsto mn(M \cap N)$. It is a surjective homomorphism since $G = MN$ and the kernel is $(M \cap N)$, so by the First Isomorphism Theorem, the statement holds. $\blacksquare$
Proposition 21 (Cauchy’s Theorem): If $G$ is a finite abelian group and $p$ is a prime dividing $\vert G \vert$, then $G$ contains an element of order $p$.
Proof: By induction. Let $G$ be a finite abelian group of order $n > 1$. Suppose the statement holds for all such groups with order less than $n$. Since $n > 1$, there is an $x \in G$ with $x \neq 1$. If $n = p$ then $\vert x \vert = p$ by Lagrange’s and we are done.
So suppose $n > p$. Choose $x \in G$ and define $H = \langle x \rangle$ with $m = \vert H \vert$. If $p$ divides $m$ then $m = kp$ and $\vert x^k \vert = p$. So suppose $p$ does not divide $m$. $m$ divides $n$ by Lagrange’s and is normal in $G$ since it is abelian. So $G/H$ is defined and $\vert G/H \vert = n/m$. Since $p$ does not divide $m$ and it divides $n$, it divides $n/m$ so $G/H$ contains an element, call it $yN$, of order $p$ by the inductive hypothesis. Since $\vert yN \vert \neq 1$ we have $y \not\in N$, but $y^p \in N$ since $(yN)^p = 1$. We have $\langle y^p \rangle \leq \langle y \rangle$ so $\vert y^p \vert \leq \vert y \vert$. Also $\langle y^p \rangle \leq N$ but $\langle y \rangle \nleq N$, so $\langle y^p \rangle \neq \langle y \rangle$ so $\vert y^p \vert \neq \vert y \vert$. So $\vert y^p \vert \lt \vert y \vert$. But by Proposition 2.5, $\vert y^p \vert = \vert y \vert/(\vert y \vert, p)$. $\vert y^p \vert \lt \vert y \vert$ so $(\vert y \vert, p) > 1$, so $p \mid \vert y \vert$ since $p$ is prime. So $lp = \vert y \vert$ for some integer $l$, so $y^l$ has order $p$ in $G$. $\blacksquare$
Definition: A group $G$ is called simple if $\vert G \vert \gt 1$ and the only normal subgroups of $G$ are $1$ and $G$.
Definition: In a group $G$, a sequence of subgroups
\[1 = N_0 \leq N_1 \leq \dots \leq N_k = G\]is called a composition series if $N_i \trianglelefteq N_{i + 1}$ and $N_{i + 1}/N_i$ is a simple group for $0 \leq i \leq k - 1$. Then, the quotient groups $N_{i + 1}/N_i$ are called composition factors of $G$.
Theorem 22 (Jordan-Holder): Let $G$ be a finite, non-trivial group. Then
Proof: $(1)$ We induct on $\vert G \vert = n$. The base case of $n = 2$ holds since if $n = 2$ then $G = Z_2$ and $1 \leq Z_2$ is a composition series for $Z_2$. For the inductive step, suppose $n > 2$. If $G$ has no nontrivial proper normal subgroups, then $1 \leq G$ is a composition series so suppose otherwise. Let $N$ be such a normal subgroup. Then $\vert G/N \vert \leq \vert G \vert$ so by the inductive hypothesis, $G/N = \overline G$ has a composition series $\overline{G_1}, \dots, \overline{G_N}$. By the Lattice Isomorphism Theorem, there is some $G_i \leq G$ for each $\overline {G_i} \leq \overline G$, and $G_i \trianglelefteq G_{i + 1}$ for all $0 \leq i \lt n$ with $G_n = G$. Finally, by the Third Isomorphism Theorem, $\overline{G_{i + 1}}/\overline{G_i} \cong G_{i + 1}/G_i$, so each composition factor is simple. So $G_1, \dots, G_n$ is a composition series for $G$.
$(2)$ TODO.
Theorem: There are 18 (infinite) families of simple groups and 26 simple groups not belonging to these families, called the sporadic simple group, such that every finite simple group is isomorphic to one of the groups in the list.
Theorem (Feit-Thompson): If $G$ is a simple group of odd order, then $G \cong Z_p$ for some prime $p$.
Proof: Out of scope (understatement).
Definition: A group $G$ is solvable if there is a chain of subgroups
\[1 = G_0 \trianglelefteq G_1 \trianglelefteq \dots \trianglelefteq G_s = G\]such that $G_{i + 1}/G_i$ is abelian for $0 \leq i \lt s$.
Proposition: If $N$ and $G/N$ are solvable, then $G$ is solvable.
Proof: Let $\overline G = G/N$ and $1 = N_0 \trianglelefteq N_1 \dots \trianglelefteq N_n = N$ and $1 = \overline G_0 \trianglelefteq \overline G_1 \trianglelefteq \dots \trianglelefteq \overline G_m = \overline G$ be the composition series with abelian factors for $N$ and $\overline G$, respectively. By the Fourth Isomorphism Theorem, $\overline G_i \cong G_i/N$ for some subgroups $G_i$ of $G$ with $N \trianglelefteq G$ and $G_i \trianglelefteq G_{i + 1}$ for $0 \leq i \lt m$. By the Third Isomorphism Theorem, $\overline G_{i + 1}/\overline G \cong G_{i + 1}/G_i$ so $G_{i + 1}/G_i$ is abelian. Thus
\[1 = N_0 \trianglelefteq N_1 \trianglelefteq \dots \trianglelefteq N_n = N = G_0 \trianglelefteq G_1 \trianglelefteq \dots \trianglelefteq G_m = G\]is a composition series with abelian composition factors. Thus $G$ is solvable. $\blacksquare$
Many important properties are shown in the exercises, such as:
Exercise 1: Prove that if $G$ is an abelian simple group then $G \cong Z_p$ for some prime $p$ ($G$ is not necessarily finite).
Proof: If $G$ is an abelian simple group, then for $\langle x \rangle \leq G$ for $x \in G - \lbrace 1 \rbrace$ we have $\langle x \rangle = G$, since otherwise $\langle x \rangle$ would be a normal proper subgroup of $G$; a contradiction. So $G$ is cyclic and $G \cong Z_n$ for some $n$. If $n$ were not prime, then $\langle d \rangle$ is a proper subgroup of $Z_n$; a contradiction. So $G \cong Z_p$ for some prime $p$. $\blacksquare$
Exercise 2: Exhibit all 3 composition series for $Q_8$ and all 7 composition series for $D_8$. List the composition factors in each case.
The composition series for each is every path from 1 to $Q_8$ (or $D_8$) in the lattice structure. For $D_8$:
| Composition Series | Composition Factors |
|---|---|
| $1, \langle r^2 \rangle, \langle r \rangle, D_8$ | $Z_2, Z_2, Z_2$ |
| $1, \langle s \rangle, D_4, D_8$ | as above |
| $1, \langle r^2s \rangle, D_4, D_8$ | as above |
| $1, \langle r^2 \rangle, D_4, D_8$ | as above |
| $1, \langle r^2 \rangle, D_4’, D_8$ | as above |
| $\dots$ | $\dots$ |
Exercise 4: Use Cauchy’s Theorem and induction to show that a finite abelian group has a subgroup of order $d$ for each positive divisor $d$ of its order.
Proof: By induction. The base case is when $\vert G \vert = 1$ which is trivial. For the inductive step, suppose $\vert G \vert = n$. Let $d$ be a divisor of $n$. $d = kp$ for some prime $p$, so $p \mid n$ and by Cauchy’s Theorem there exists a subgroup $H \leq G$ of order $p$. Since $G$ is abelian, $H$ is normal, so take $G/H$ with order $n/p$. By the inductive hypothesis, let $K/H \leq G/H$ with order $d/p$. By the Correspondence Theorem, $\vert K \vert = \vert H \vert d/p = d$. So $K \leq G$ with order $d$. $\blacksquare$
Exercise 5: Prove that subgroups and quotient groups of a solvable group are solvable.
Proof: Let $G$ be solvable with a series $G_1, \dots G_n$ and $H \leq G$. We show that $H$ is solvable with series $H_i = G_i \cap H$, $0 \leq i \leq n$. Note that $H_{i} = G_i \cap H = (G_{i + 1} \cap H) \cap G_i$. So by the Second Isomorphism Theorem,
\[\frac{H_{i + 1}}{H_i} = \frac{G_{i + 1} \cap H}{(G_{i + 1} \cap H) \cap G_i} \cong \frac{(G_{i + 1} \cap H)G_i}{G_i}\]so $H_{i} \trianglelefteq H_{i + 1}$. Also, $(G_{i + 1} \cap H)G_i \leq G_{i + 1}$, so by the Lattice Isomorphism Theorem, $\frac{(G_{i + 1} \cap H)G_i}{G_i} \leq G_{i + 1}/G_i$. so $H_{i + 1}/H_i$ is abelian.
For the quotient group of a solvable group, let $G$ be solvable with series $G_1, \dots, G_n$ and $N \trianglelefteq G$. Define $\overline G = G/N$. $\overline G$ is solvable with series $\overline G_1, \dots, \overline G_n$ since by the Lattice Isomorphism Theorem, $\overline{G_{i}} \trianglelefteq \overline{G_{i + 1}}$ since $G_{i} \trianglelefteq G_{i + 1}$, and $\overline{ G_{i + 1}} / \overline {G_i}$ is abelian since by the Third Isomorphism Theorem, $\overline{ G_{i + 1}}/\overline {G_i} \cong G_{i + 1}/G_i$. $\blacksquare$
Exercise 6: Proof of $(1)$ of Jordan-Holder.
Exercise 7: If $G$ is a finite group and $H \trianglelefteq G$, prove that there is a composition series of $G$, one of whose terms is $H$.
Proof: By $(1)$ of Jordan-Holder, $H$ has a composition series $1 = H_1 \trianglelefteq \dots \trianglelefteq H_n = H$. Since $H \trianglelefteq G$, $\overline G = G/H$ is well-defined and also has a composition series $1 = \overline{G_1} \trianglelefteq \dots \trianglelefteq \overline{G_m} = \overline G$. By the Third and Fourth Isomorphism Theorems, there is a $G_i \leq G$ such that $\overline G_i = G_i/N$ and $H = G_1 \trianglelefteq \dots \trianglelefteq G_m = G$ is a composition series from $H$ to $G$ (see the proof of $(1)$ of Jordan-Holder). So
\[1 = H_1 \trianglelefteq \dots \trianglelefteq H_n = H = G_1 \trianglelefteq \dots \trianglelefteq G_m = G\]is a composition series for $G$, one of whose terms is $H$. $\blacksquare$
Exercise 8: Let $G$ be a finite group. Prove that the following are equivalent.
Proof: $(1 \to 2)$ Suppose $G$ is solvable with series $1 = H_0 \trianglelefteq \dots \trianglelefteq H_s = G$ where $s$ is maximal. Suppose there is a factor $H_{i + 1}/H_i$ that is non-cyclic. Then, it has a cyclic proper subgroup $X/H_i$. Since the factors are abelian, $X/H_i \trianglelefteq H_{i + 1}/H_i$, so by the Lattice Isormorphism Theorem, $X \trianglelefteq H_{i + 1}$. So $X$ can be placed in between $H_i$ and $H_{i + 1}$ in the series, contradicting that $s$ is maximal. So every factor is cyclic.
$(2 \to 3)$ Let $1 = H_0 \trianglelefteq \dots \trianglelefteq H_s = G$ be the subnormal series existing by the assumption with maximal $s$. Suppose there is a quotient $H/N$ that is not of prime order. Since it is cyclic, there is a subgroup $M/N \trianglelefteq H/N$, so $N \trianglelefteq M \trianglelefteq H$, so $M$ may be added to the subnormal series between $N$ and $H$, contradicting that $s$ is maximal. So every quotient is of prime order. Every quotient is prime and cyclic, so it is simple, so the series is a composition series with factors of prime order. By Jordan-Holder, this composition series is unique, so all composition factors are of prime order.
$(3 \to 4)$ TODO.
$(4 \to 1)$ TODO.
Definition: A 2-cycle is called a transposition.
Proposition: Every element of $S_n$ may be written as a product of transpositions.
Proof: Note that $(a_1a_2 \dots a_m) = (a_1a_m)(a_1a_{m - 1})\dots(a_1a_2)$ for any $m$-cycle, so we may write any element in $S_n$ as a product of cycles (its cycle decomposition), and then each of these cycles in turn as a product of transpositions as in the above. $\blacksquare$
Example: The product of transpositions for $\sigma = (1,12,8,10,4)(2,13)(5,11,7)(6,9)$ is
\[\sigma = (1,4)(1,10)(1,8)(1,12)(2,13)(5,7)(5,11)(6,9)\]Definition: Let $x_1, \dots, x_n$ be independent variables and
\[\Delta = \prod_{1 \leq i \lt j \leq n}(x_i - x_j)\]and let $\sigma \in S_n$ act on $\Delta$ by permuting the variables by its indices as
\[\sigma(\Delta) = \prod_{1 \leq i \lt j \leq n} \left(x_{\sigma(i)} - x_{\sigma(j)}\right).\]Now, define
\[\epsilon(\sigma) = \begin{cases} 1, &\sigma(\Delta) = \Delta \\ -1, &\sigma(\Delta) = -\Delta \end{cases}\]Then, $\epsilon(\sigma)$ is the sign of $\sigma$ and $\sigma$ is an even permutation if $\epsilon(\sigma) = 1$ and and odd permutation if $\epsilon(\sigma) = -1$.
Proposition 23: The map $\epsilon: S_n \to \lbrace \pm 1 \rbrace$ is a homomorphism.
Proof: Let $\sigma, \tau \in S_n$. Note that $\sigma(\Delta) = \epsilon(\sigma)\Delta$ and similarily $\tau(\Delta) = \epsilon(\tau)\Delta$. Then
\[(\tau \sigma)(\Delta) = \epsilon(\sigma)\tau(\Delta)=\epsilon(\sigma)\epsilon(\tau)\Delta\]Thus $(\tau\sigma)(\Delta) = \epsilon(\sigma\tau)\Delta = \epsilon(\sigma)\epsilon(\tau)\Delta$, so $\epsilon(\sigma\tau) = \epsilon(\sigma)\epsilon(\tau)$. $\blacksquare$
Proposition 25: Transpositions are all odd permutations and $\epsilon$ is a surjctiv ehomomorphism.
Proof: Let $(a, b) \in S_n$, $1 \leq a,b \leq n$ be a transposition. Let $\lambda = (1,a)(2, b)$. Then $(a, b) = \lambda(1,2)\lambda$ so
\[\begin{align*} \epsilon((a, b)) &= \epsilon(\lambda(1, 2)\lambda) &= \epsilon(\lambda)^2 \epsilon((1, 2)) \\ &= \epsilon((1, 2)) \\ &= -1 \end{align*}\]$\blacksquare$
Definition: The alternating group of degree $n$, denoted by $A_n$, is the kernel of $\epsilon$, that is, the set of even permutations.
Remark: The product of two permutations is even if and only if either both permutations are even or both are odd.
Proposition 25: The permutation of $\sigma$ is odd if and only if the number of cycles of even length in its cycle decomposition is odd.
Proof: A cycle of even length has an odd number of transpositions, so the cycle is odd. $(-1)^k = -1$ if and only if $k$ is odd. $\blacksquare$
Some useful tools used in the Exercises:
Exercise 2: Prove that $\sigma^2$ is an even permutation for every permutation $\sigma$.
Proof: $\epsilon(\sigma^2) = \epsilon(\sigma)\epsilon(\sigma) = 1$ for all $\epsilon(\sigma) \pm 1$. $\blacksquare$
Exercise 3: Prove that $S_n$ is generated by $\lbrace (i,i+1) \mid 1 \leq i \leq n - 1 \rbrace$.
Proof: We know that $S_n$ may be generated by any transposition. It suffices to show that the adjacent transpositions generate the set of transpositions. It is straightforward to show by induction that $(1, k) = (k,k-1)(1,k-1)(k,k-1)^{-1}$ for all $2 \leq k \leq n$, and similarily for $(2, k)$, $(3, k)$, etc. $\blacksquare$
Exercise 4: Show that $S_n = \langle (1,2), (1,2,3,\dots,n) \rangle$
Proof: $(1,2,\dots,n)$ can be represented as a product of adjacent transpositions $(1,2)(2,3)\dots(n - 1, n)$, from which the result follows from Exercise 3.
Exercise 5: Show that if $p$ is prime, then $S_p = \langle \sigma, \tau \rangle$ where $\sigma$ is any transposition and $\tau$ is any $p$-cycle.
Proof: Without loss of generality, let $\tau = (1, 2, \dots, p)$ and $\sigma = (a, b), 1 \leq a \lt b \leq p$. Let $d = b - a$. We may produce all $p$ transpositions of the form $(i, (i + d) \bmod p)$ with $\tau^{-i}\sigma\tau^i$ for all $0 \leq i \leq p$. Then, given two transpositions $\alpha = (1, (1 + d) \bmod p)$ and $\beta = ((1 + d) \bmod p, (1 + 2d) \bmod p)$, one has $\alpha_2 = (1, (1 + 2d) \bmod p) = \alpha^{-1}\beta\alpha$, and in general, $\alpha_k = (1, (1 + kd) \bmod p)$. By Bezout’s Theorem, $kd \bmod p = 1$ for some $k$, so we have $(1, 2)$. Then, by Exercise 4, the statement follows. $\blacksquare$
Exercise 6: Show that $\langle (1\enspace 3), (1 \enspace 2 \enspace 3 \enspace 4) \rangle$ is a proper subgroup of $S_4$. What is the isomorphism type of this subgroup?
Let $\alpha = (1 \enspace 3)$ and $\beta = (1 \enspace 2 \enspace 3 \enspace 4)$. Then the subgroup generated is $\lbrace 1, \beta, \beta^2, \beta^3, \alpha\beta, \alpha\beta^2, \alpha\beta^3 \rbrace$, which is isomorphic to $D_8$.