Problem: Let $G = D_4 \times \mathbb{Z}_3$. Prove that $G \space / \space Z(G)$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$.
Here $Z(G)$ is the center of G, which I have figured out is of order 6. So, $|G \space / \space Z(G)|=24\space/\space6=4$ so whatever it is isomorphic to must have order 4. Now since G is not abelian then $G \space / \space Z(G)$ is not cyclic and cannot be isomorphic to $\mathbb{Z}_4$ (is this true? I think it is). So, I can see it must instead be isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$.
But how do I prove this rigorously? Or, is there an isomorphism that takes $G \space / \space Z(G)$ to $\mathbb{Z}_2 \times \mathbb{Z}_2$ that is easy to find? I haven't been able to come up with one yet.
Any help would be appreciated!