# Show that $S_4$ is not isomorphic to $D_4\times\mathbb{Z}_3$

Show that $S_4$ is not isomorphic to $D_4\times\mathbb{Z}_3$

I have no idea how to show this. I'm studying for a test, so I am less interested in solutions and hints than I am strategy. What should my thought process be as I try to solve this problem? What kind of thing do I look for?

• Look for properties of one abstract group not shared with the other. For example, the number of elements of given order. – David Feb 9 '16 at 18:26
• Fact: if $n > 3$, and $\sigma \in S_n$ commutes with every other element of $S_n$, then $\sigma$ is the identity. In other words, the center of $S_n$ is trivial. Is the center of $D_4 \times Z_3$ trivial? – D_S Feb 9 '16 at 19:03

For example, $D_4\times\mathbb{Z}_3$ has an element of order $3$ in its center. The elements of order $3$ in $S_4$ are the $3$-cycles, none of which are in the center.
For another example, $D_4\times \mathbb{Z}_3$ has an element of order $6$. $S_4$ has no elements of order $6$.
For yet another example, $D_4\times\mathbb{Z}_3$ has $5$ elements of order $2$. $S_4$ has $(12),(13),(14),(23),(24),(34),\ldots$ which is already too many.