TITLE: Prove $D_{24}$ is not isomorphic to $S_4$.
QUESTION [7 upvotes]: The question from "Abstract Algebra," by Dummit & Foote is: 

Prove that the dihedral group $D_{24}$ is not isomorphic to $S_4$.

We can express the presentation of 
$$D_{24}=\{r,s | r^{12}=s^2=1, rs=sr^{-1}\}.$$
Obviously for some permutation $\sigma \in S_4, \sigma ^4=1$ and $(1 2)(1 2)=1$. So $S_4$ can also be expressed as $$S_4=\{ \sigma, k | \sigma^4=k^2=1\},$$ where $k=(1 2)$ so $D_{24}$ is homomorphic to $S_4$.
Also, $|D_{24}|=24$ and $S_4=4!$ so $D_{24} \cong S_4$. This is contradicting the question.
Why am I wrong?
Thanks in advance!

REPLY [5 votes]: $D_{24}$ and $S_4$ are not isomorphic,  because $S_4$ has trivial center, but $D_{24}$ does not - see this question: Center of dihedral group
Two finite groups (of the same order) being "homomorphic" does not mean they are "isomorphic".