Group Theory
Finite Groups
Let $G$ be a finite group.
- If for all positive integers $n$ dividing its order, $G$ contains at most $n$ elements $x$ satisfying $x^n = 1$, then $G$ is cyclic.
- Proof: Show $G$ is nilpotent, and each of its Sylow subgroups are cyclic.
- If $H \trianglelefteq G$ and $P \in Syl_p(H)$, then $G = HN_G(P)$ and $\vert G : H \vert \mid \vert N_G(P) \vert$.
- Proof: Show $G \leq HN_G(P)$ since $P^g = P^h$. Use Second Isomorphism Theorem for the second part.
- If $\vert G \vert = p^aq^b$ for primes $p, q$, then $G$ is solvable.
- If for every prime $p$ dividing $\vert G \vert$, $\vert G \vert = p^\alpha m$, $G$ has a subgroup of order $m$, then $G$ is solvable.
- If $\vert G \vert$ is odd then $G$ is solvable.
- If for every pair of elements $x, y \in G$, $\langle x, y \rangle$ is a solvable group.
Finite Abelian Groups
Let $A$ be a finite abelian group.
- If $\vert A \vert$ is the product of distinct primes, then up to isomorphism the only abelian group of order $n$ is $Z_n$.
- Proof: Use the Classification Theorem for Finitely Generated Abelian Groups.
- $A$ is direct product of its Sylow subgroups.
- Proof: Use finite nilpotent groups.
Direct Products
Let $G = G_1 \times \dots \times G_n$. Identify $G_i$ with $\lbrace (1, \dots, g_i \dots, 1) \mid g_i \in G_i \rbrace$ when necesssary.
- The order of a direct product is the product of the orders of its factors.
- $G/G_i \cong G_1 \times \dots G_{i - 1} \times G_{i + 1} \times G_n$.
- The projection of $G \to G_i$ is a surjective homomorphism.
- If $x \in G_i$ and $y \in G_j$ then $xy = yx$ for $i \neq j$.
Finitely Generated Abelian Groups
Let $A$ be a finitely generated abelian group.
- $A \cong \mathbb Z^r \times Z_{n_1} \times \dots \times Z_{n_s}$ is unique with $n_s \mid \dots \mid n_1$.
- If $n$ is the product of distinct primes, then up to isomorphism the only abelian group of order $n$ is $Z_n$.
Commutator Subgroups
Let $G’$ be the commutator subgroup of $G$.
- The homomorphic image of a commutator subgroup is a commutator subgroup.
- $G’$ char $G$.
- $G/G’$ is abelian.
- $G/G’$ is the largest abelian quotient of $G$: $G/H$ is abelian if and only if $G’ \leq H$.
p-groups
Let $P$ be a $p$-group of order $p^\alpha$.
- $Z(P) \neq 1$.
- Proof: Use class equation.
- proper normal subgroups of $P$ intersect the center non-trivially.
- Proof: Normal subgroups are partitioned by conjugacy classes + class equation.
- normal subgroups $H$ of $P$ contain normal subgroups of order $p^b$ for every divisor $p^b$ of $\vert H \vert$
- Proof: By induction on $a$, use Cauchy’s Theorem to generate a normal subgroup $N \trianglelefteq H \cap Z(P)$ and induct on $P/N$.
- proper subgroups of $P$ are contained in their normalizer.
- Proof: By induction on $\vert P \vert$, induct on $P/Z(P)$.
- every maximal subgroup of $P$ has index $p$ and is thus normal in $P$
- Proof: Show $P/M$ contains no proper nontrivial subgroups, so by $(3)$ $\vert P/M \vert = p$
- $P$ is nilpotent of class at most $a - 1$.
- Proof: Show $\vert H_i \vert \geq p\vert H_{i - 1} \vert$ since $Z(P/H_{i - 1}) \neq 1$.
Examples
- $Z_2, Z_4, Z_8, D_8, D_{16}, D_{32}, P \in Syl_2(S_3)$
Nilpotent Groups
Finite Nilpotent Groups
Let $G$ be a finite nilpotent group with distinct primes $p_i$ dividing $\vert G \vert$ and corresponding Sylow-$p_i$ groups $P_i$.
- The following are equivalent
- $G$ is nilpotent.
- If $H \lt G$ then $H \lt N_G(H)$
- Proof: (i $\to$ ii) Same argument as for $p$-groups, using the fact that if $G$ is nilpotent then so is $G/Z(G)$.
- $G$’s Sylow groups are normal in $G$.
- $G$ is the direct product of its Sylow groups.
- Every maximal subgroup is normal.
- $G^n = 1$ for some $n \geq 0$. $G$ is of class $c$ iff $c$ is the smallest non-negative integer with $G^c = 1$.
Examples
Solvable Groups
Let $G$ be a solvable group.
- The following are equivalent.
- $G$ is solvable.
- $G^{(n)} = 1$ for some $n \geq 0$.
- Subgroups of solvable groups are solvable.
- Proof: Show $H^{(i)} \leq G^{(i)}$ for $H \leq G$.
- Homomorphic images and quotient groups of solvable groups are solvable.
- Proof: Show $\varphi(G^{(i)}) = K^{(i)}$.
- If $N$ is normal in $G$ and both $N$ and $G/N$ are solvable then so is $G$.
- Proof: Use the above property with the natural projection.