Question: > 
> Ιs $A\_4 \times \mathbb{Z}\_3$ isomorphic to $S\_3 \times S\_3$?
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> 
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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: 
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 $\mathbb{Z}\_2\times\mathbb{Z}\_2\times\mathbb{Z}\_3$.
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 $n \geq 3$.