Definition: The direct product $G_1 \times G_2 \times \dots$ for groups $G_1, G_2, \dots$ with operations $\star_1, \star_2, \dots$, respectively, is the set of $n$-tuples $(g_1, \dots, g_n)$ (if finite) or sequences (if infinite) with the operation defined componentwise.
Proposition 1: If $G_1, \dots, G_n$ are groups, their direct product is a group of order $\prod \vert G_i \vert$ (if any $G_i$ is infinite, so is the direct product).
Proof: Trivial.
Proposition 2: Let $G_1, \dots, G_n$ be groups and $G$ their direct product.
Proof: $(1)$ The map for the first isomorphism is given by $g_i \mapsto (1, \dots, 1, g_i, 1, \dots, 1)$. For the second part, note that
\[\varphi: G \to G_1 \times \dots \times G_{i -1} \times G_{i + 1} \times \dots \times G_n\]defined by $(\varphi(g_1, \dots, g_n)) = (g_1, \dots, g_{i - 1}, g_{i + 1}, \dots, g_n)$ is a homomorphism with kernel $G_i$.
$(2)$ It is clear that this map is a surjective homomorphism: to see the kernel note that $G/G_i \cong \lbrace (g_1, \dots, g_{i - 1}, 1, g_{i + 1}, \dots, g_n) \mid g_j \in G_j \forall j \neq i \rbrace$.
$(3)$ Trivial.
$\blacksquare$
Definition: $G$ is finitely generated if there exists a finite $X \subseteq G$ such that $G = \langle X \rangle$.
Definition: Let $r \in \mathbb Z_{\geq 0}$. Then $\mathbb Z^r = \mathbb Z \times \dots \times \mathbb Z$ ($r$ times) is the free Abelian group of rank $r$. When $r = 0$, $\mathbb Z^0 = 1$.
Examples:
Theorem 3 (Fundamental Theorem of Finitely Generated Abelian Groups, Invariant Form): Let $G$ be a finitely generated Abelian group. Then
Proof: Ring theory. $\blacksquare$
Definition: The exponent $r$ in the $\mathbb Z^r$ term of the invariant decomposition is called the Betti number of $G$ or the free rank of $G$, and $n_1, \dots, n_s$ are the invariants of $G$.
Remarks:
Corollary 4: If $n$ is the product of distinct primes, then up to isomorphism the only abelian group of order $n$ is the cyclic group of order $n$, $Z_n$
Proof: By Theorem 3, since $n_1 \mid n_2 \mid \dots$ and $n$ is the product of its invariant factors, any prime divisor of $n$ must divide $n_1$. But $n_1$ is prime, so $n_1 = n$, so by Theorem 3, $G \cong Z_n$. $\blacksquare$
Remark: $G$ is an Abelian group if its free rank is 0.
Theorem 5: Let $G$ be an abelian group of order $n \gt 1$ with unique prime factorization $n = p_1^{\alpha_1}\dots p_k^{\alpha_k}$. Then
with $\beta_1 \geq \dots \geq \beta_t \geq 1$ and $\sum \beta = \alpha$.
Definition: The integers $p^{\beta_j}$ in Theorem 5 are the elementary divisors of $G$. The decomposition in the same theory is the elementary divisor decomposition of $G$.
Remark: Theorem 5 states that finite abelian groups are isomorphic to the direct product of its Sylow subgroups.
Example: Listing all abelian groups of order $n$.
$n = 1800 = 2^33^25^2$, so we have
Proposition 6: Let $m, n \in \mathbb Z^+$. Then
If $G$ is abelian with type $(n_1, \dots, n_s)$ where each $n_i$ has prime factor decomposition $n_i = p_1^{\beta_{i1}} \dots p_k^{\beta_{ik}}$, then the elementary divisors of $G$ are precisely the $p_j^{\beta_{jt}}$ factors.
This process also works if $G$ is expressed as its cyclic decomposition, e.g. $G \cong Z_6 \times Z_{15}$ (note that these aren’t invariant factors since $6 \nmid 15$).
Create a non-increasing list of prime powers for each prime $p$. Make these lists equal by adding “padding” of 1’s. Then, form the invariant factors by taking the product of the $i$th integer in each of the lists.
For example, for $G$ with elementary divisors 2, 3, 2, 25, 3, form the lists
| $p = 2$ | $p = 3$ | $p = 5$ | |
|---|---|---|---|
| 2 | 3 | 25 | |
| 2 | 3 | 1 | |
| 2 | 1 | 1 |
then the invariant factors are $2 \cdot 3 \cdot 25$, $2 \cdot 3 \cdot 1$, $2 \cdot 1 \cdot 1$, so
\[G \cong Z_{150} \times Z_6 \times Z_2\]Definition:
Definition: Let $G$ be a group, $x, y \in G$, and $A, B$ non-empty subsetes of $G$.
Proposition 7: Let $G$ be a group, $x, y \in G$ and $H \leq G$.
Proof: $(1)$ By definition.
$(2)$ If $H \trianglelefteq G$, then $h^{-1}g^{-1}hg \in h^{-1}H = H$, so $[H, G] \leq H$. Conversely if $h^{-1}g^{-1}hg \in H$, then $g^{-1}hg \in H$ so $H$ is normal.
$(3)$ Let $x, y \in G$. Then $\sigma[x, y] = \sigma(x^{-1}y^{-1}xy) = \sigma(x)^{-1}\sigma(y)^{-1}\sigma(x)\sigma(y) = [\sigma(x), \sigma(y)]$. So $\sigma$ maps the commutators of $G$ bijectively onto itself. Since the commutators are a generating set of $G’$, $\sigma(G’) = G’$, that is, $G’$ char $G$.
Finally, we show $G/G’$ is abelian. Let $xG’, yG’ \in G/G’$. Then $(xG’)(yG’) = (xy)G’ = (yx[x, y])G’ = (yx)G’$.
$(4)$ Suppose $H \trianglelefteq G$ with $G/H$ abelian. Then $(xy)H = (yx)H$ for $x, y \in G$ so $[x, y]H = H$, so $G’ \leq H$. Conversely if $G’ \leq H$, then since $G/G’$ is abelian by $(3)$, every subgroup of $G/G’$ is normal, including $H/G’$. So by the Lattice Isomorphism Theorem, $H \trianglelefteq G$. And by the Third Isomorphism Theorem, $G/H \cong (G/G’)/(H/G’)$, so $G/H$ is abelian.
$(5)$ Let $H = \operatorname{ker} \varphi$. Then $G’ \leq H$ by $(4)$. TODO.
$\blacksquare$
Proposition 8: Let $H, K \leq G$. The number of distinct ways of writing each element of $HK$ in the form $hk$ for some $h \in H, k \in K$ is $\vert H \cap K \vert$.
Proof: Let $x \in HK$. Since $HK = \bigcup_{h \in H} hK$ and the cosets of $K$ partition $HK$, $x$ is in some unique coset $hK$. The number of ways of writing $