Groups: Classification

Spring 2020 #1 #algebra/qual/work

\todo[inline]{Work this problem.}

How many isomorphism classes are there of groups of order 45?

Describe a representative from each class.

concept:

solution:

It turns out that $n_3 = 1$ and $n_5 = 1$ , so $G \cong S_3 \times S_5$ since both subgroups are normal.
There is only one possibility for $S_5$ , namely $S_5\cong {\mathbf{Z}}/(5)$ .
There are two possibilities for $S_3$ , namely $S_3 \cong {\mathbf{Z}}/(3^2)$ and ${\mathbf{Z}}/(3)^2$ .
Thus
$G \cong {\mathbf{Z}}/(9) \times{\mathbf{Z}}/(5)$ , or
$G \cong {\mathbf{Z}}/(3)^2 \times{\mathbf{Z}}/(5)$ .

\todo[inline]{Revisit, seems short.}

Let $G$ be a group of order 70.

Show that $G$ is not simple.
Exhibit 3 nonisomorphic groups of order 70 and prove that they are not isomorphic.

How many groups are there up to isomorphism of order $pq$ where $p<q$ are prime integers?

Use the Class Equation (equivalently, the conjugation action of a group on itself) to prove that any $p{\hbox{-}}$ group (a group whose order is a positive power of a prime integer $p$ ) has a nontrivial center.
Prove that any group of order $p^2$ (where $p$ is prime) is abelian.

Prove that any group of order $5^2 \cdot 7^2$ is abelian.
Write down exactly one representative in each isomorphism class of groups of order $5^2 \cdot 7^2$ .

concept:

Centralizer: $C_G(x) = \left\{{g\in G {~\mathrel{\Big\vert}~}[gx] = 1}\right\}$ .
Class Equation: ${\left\lvert {G} \right\rvert} = {\left\lvert {Z(G)} \right\rvert} + \sum [G: C_G(x_i)]$
$G/Z(G)$ cyclic $\iff G$ is abelian. $\begin{align*} G/Z(G) = \left\langle{xZ}\right\rangle &\iff g\in G \implies gZ = x^mZ \\ &\iff g(x^m)^{-1}\in Z \\ &\iff g = x^m z {\quad \operatorname{for some} \quad}z\in Z\\ &\implies gh = x^mz_1 x^n z_2 = x^n z_2 x^m z_1 = hg .\end{align*}$
Every group of order $p^2$ is abelian.
Classification of finite abelian groups.

solution:

proof (of a):

Strategy: get $p$ to divide ${\left\lvert {Z(G)} \right\rvert}$ .

Apply the class equation: $\begin{align*} {\left\lvert {G} \right\rvert} = {\left\lvert {Z(G)} \right\rvert} + \sum [G: C_G(x_i)] .\end{align*}$
Since $C_G(x_i) \leq G$ and ${\left\lvert {G} \right\rvert} = p^k$ , by Lagrange ${\left\lvert {C_G(x_i)} \right\rvert} = p^\ell$ for some $0\leq \ell \leq k$ .
Since ${\left\lvert {G} \right\rvert} = p^k$ for some $k$ and $Z(G), C_G(x_i) \leq G$ are subgroups, their orders are powers of $p$ .
Use $\begin{align*}[G: C_G(x_i)] = 1 \iff C_G(x_i) = G \iff \left\{{g\in G{~\mathrel{\Big\vert}~}gx_ig^{-1}= x_i}\right\} = G \iff x_i \in Z(G).\end{align*}$
- Thus every index appearing in the sum is greater than 1, and thus equal to $p^{\ell_i}$ for some $1\leq \ell_i \leq k$
- So $p$ divides every term in the sum
Rearrange $\begin{align*} {\left\lvert {G} \right\rvert} - \sum [G: C_G(x_i)] = {\left\lvert {Z(G)} \right\rvert} .\end{align*}$
$p$ divides both terms on the LHS, so must divide the RHS, so ${\left\lvert {Z(G)} \right\rvert} \geq p$ .

proof (of b):

Strategy: examine ${\left\lvert {G/Z(G)} \right\rvert}$ by cases.

proof (of c):

By Sylow
- $n_5 \divides 7^2,\quad n_5\cong 1\operatorname{mod}5 \implies n_5\in\left\{{1, 7, 49}\right\}\setminus\left\{{7, 49}\right\} = \left\{{1}\right\} \implies n_5 = 1$
- $n_7 \divides 5^2, \quad n_7 \cong 1 \operatorname{mod}7 \implies n_7 \in \left\{{1, 5, 25}\right\}\setminus\left\{{5, 25}\right\} =\left\{{1}\right\} \implies n_7 = 1$
By recognition of direct products, $G = S_5 \times S_7$
- By above, $S_5, S_7{~\trianglelefteq~}G$
- Check $S_5\cap S_7 = \left\{{e}\right\}$ since they have coprime order.
- Check $S_5S_7 = G$ since ${\left\lvert {S_5 S_7} \right\rvert} = 5^2 7^2 = {\left\lvert {G} \right\rvert}$
By (b), $S_5, S_7$ are abelian since they are groups of order $p^2$
The direct product of abelian groups is abelian.

proof (of d):