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.

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 \(5^2 \cdot 7^2\).

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.

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.