Is $A_4 \times \mathbb{Z}_3$ isomorphic to $S_3 \times S_3$

abstract-algebrafinite-groupsgroup-theorysymmetric-groups

Ιs $A_4 \times \mathbb{Z}_3$ isomorphic to $S_3 \times S_3$?

I am trying to find an element of $S_3 \times S_3$ which has an order, let's say $a$, and $A_4 \times \mathbb{Z}_3$ has no element of such order.

Best Answer

In $S_3\times S_3$ there are

  1. $1$ element of order $1$
  2. $\color{red}{15}$ elements of order $2$, $(\tau_1,e),(e,\tau_2),(\tau_1,\tau_2)$
  3. $8$ elements of order $3$
  4. $12$ elements of order $6$

while in $A_4\times\mathbb{Z}_3$ there are

  1. $1$ element of order $1$
  2. $\color{red}{3}$ elements of order $2$, $(\tau_1 \tau_2,e)$
  3. $26$ elements of order $3$
  4. $6$ elements of order $6$

so the easiest way to state $S_3\times S_3 \not\simeq A_4\times\mathbb{Z}_3$ is probably to compare the number of involutions (elements with order $2$) in both groups. Or wonder about the largest abelian subgroups, which in the former case are isomorphic to $\mathbb{Z}_3\times\mathbb{Z}_3$ and in the latter case are isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3$.

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