I am a beginning learner of group theory.
Which of the following are isomorphic:$$\mathbb{Z_{24}}, D_{4}\times \mathbb{Z_{3}},A_{4}\times \mathbb{Z_{2}},\mathbb{Z_{2}}\times D_{6}, \mathbb{Z_{12}}\times \mathbb{Z_{2}},D_{12}, S_{4}$$
I know $\mathbb{Z_{24}}$ is cyclic and none of $\mathbb{Z_{12}}\times \mathbb{Z_{2}},D_{12}, S_{4}$ are cyclic since none has an element of order 24. I also know $D_{12}$ and $S_{4}$ are not isomorphic because $S_{4}$ has no element of order 12. Finally, I know $A_{4}$ and $D_{6}$ are also not isomorphic because $D_{6}$, unlike $A_{4}$, contains an element of order 6.
Other than this I don't know how to proceed. Any hints are appreciated. Please allow me to do the work.
Best Answer
$\mathbb{Z_{24}}$ is the only cyclic one - all other groups are non-abelian except $\mathbb{Z_{12}}\times \mathbb{Z_{2}}$, which you noted is not cyclic.
Among the remaining ones, $\mathbb{Z_{12}}\times \mathbb{Z_{2}}$ is the only abelian one, so this is out too.
If you have already seen the center of a group, you may proceed as follows.
Among the remaining ones, $S_{4}$ is the only centerless one ($Z(S_{4}) = 1$).
Now you should know that $D_{2 k}$ has a center of order $2$. Then $D_{4}\times \mathbb{Z_{3}}$ is the only one with a center of order $6$, and $\mathbb{Z_{2}}\times D_{6}$ is the only one with a center of order $4$.
We are left with $A_{4}\times \mathbb{Z_{2}}$ and $D_{12}$. But the latter has an element of order $12$, while the former does not. In fact, if $a \in A_{4}\times \mathbb{Z_{2}}$ has order $12$, then $a^{2} \in A_{4}$ has order $6$, and you know this is not the case.