The isomorphism of $\mathbb{Z}_4\times\mathbb{Z}_2$ and symmetry group of square Are the following two groups isomorphic:
$\mathbb{Z}_4\times\mathbb{Z}_2$ and the symmetry group of the square? 
I know their orders are the same, but I don't know how to continue.
 A: No, one is abelian, and one is not.
A: No they are not isomorphic:
$\mathbb{Z}_4 \times \mathbb{Z}_2$ is abelian, so everything commutes with everything else. However, the symmetry group of the square is not abelian. For example, rotating $90^\circ$ and then flipping is not the same as flipping then rotating. Thus, since they cannot be isomoprhic. If they were by some isomoprhism $\phi$, then $\phi$ is a homomorphism so $\phi(ab) = \phi(a)\phi(b)$, and we may assume $\phi(a)$ is a flip and $\phi(b)$ is a rotation. But $ab=ba$ in $\mathbb{Z}_4\times \mathbb{Z}_2$ so $\phi(ab)=\phi(ba) = \phi(b)\phi(a)$. This shows $\phi(b)\phi(a) = \phi(a)\phi(b)$, a contradiction. 
A: As others have said, $\mathbb{Z}_4 \times \mathbb{Z}_2$ and $D_8$ (the symmetry group of a square) are not isomorphic because the former is abelian and the latter is not.
Another way to see this is to notice that $D_8$ is the semidirect product $\mathbb{Z}_4 \rtimes_\varphi \mathbb{Z}_2$ where $\varphi$ acts as inversion.  It's a theorem (in Dummit & Foote, for example) that $\mathbb{Z}_4 \rtimes_\varphi \mathbb{Z}_2$ is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$ if and only if $\varphi : \mathbb{Z}_2 \to \text{Aut}(\mathbb{Z}_4)$ is the trivial homomorphism.
