# Groups: Classification

## Spring 2020 #1#algebra/qual/work

• Show that any group of order 2020 is solvable.

• Give (without proof) a classification of all abelian groups of order 2020.

• Describe one nonabelian group of order 2020.
\todo[inline]{Work this problem.}


## Spring 2019 #3#algebra/qual/completed

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

Describe a representative from each class.

• Sylow theorems:
• $$n_p \cong 1 \operatorname{mod}p$$
• $$n_p \divides m$$.

• 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.}


## Spring 2012 #3#algebra/qual/work

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.

## Fall 2016 #3#algebra/qual/work

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

## Spring 2018 #1#algebra/qual/completed

• 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 $$5^2 \cdot 7^2$$.

• 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.

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 $${\left\lvert {Z(G)} \right\rvert} \geq p$$.

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

• $$1$$: Then $$G = Z(G)$$ and $$G$$ is abelian.
• $$p$$: Then $$G/Z(G)$$ is cyclic so $$G$$ is abelian
• $$p^2$$: Not possible, since $${\left\lvert {Z(G)} \right\rvert} > 1$$ by (a).

• 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.

• $${\mathbf{Z}}_{5^2} \times{\mathbf{Z}}_{7^2}$$
• $${\mathbf{Z}}_{5}^2 \times{\mathbf{Z}}_{7^2}$$
• $${\mathbf{Z}}_{5^2} \times{\mathbf{Z}}_{7}^2$$
• $${\mathbf{Z}}_{5}^2 \times{\mathbf{Z}}_{7}^2$$
#1 #algebra/qual/work #3 #algebra/qual/completed