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

Question

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.

Answer 1

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 $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3$.

Answer 2

You could use the following useful facts.

• Given $A$ and $B$ groups, $Z(A \times B) = Z(A) \times Z(B)$.
• $Z(S_n)=\lbrace id \rbrace$ for all $n \geq 3$.