Proposition: An abelian group is not isomorphic to an non-abelian group.
I need to prove this proposition and conclude that:
$$
\Bbb{Z}/8\Bbb{Z},\Bbb{Z}/4\Bbb{Z}\times \Bbb{Z}/2\Bbb{Z},\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}\not\cong D_4
$$
$$
\Bbb{Z}/8\Bbb{Z},\Bbb{Z}/4\Bbb{Z}\times \Bbb{Z}/2\Bbb{Z},\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}\not\cong Q
$$
($D_4$ stands for the dihedral group of order eight and $Q$ for the quaternion group of order eight)
But I have no idea how to begin? Can someone give me some hints/tips?
Best Answer
Suppose you have an abelian group $G$ and an isomorphism $\varphi: G \to H$, for some group $H$. Then for any $a,b \in G$, you have that $\varphi(ab) = \varphi(a)\varphi(b)$, but since $ab = ba$ ($G$ is abelian), we have that $\varphi(ab) = \varphi(ba) = \varphi(b)\varphi(a)$, so
$\varphi(a)\varphi(b) = \varphi(b)\varphi(a)$
And since $\varphi$ is surjective, $H$ must be abelian.
I have proved "If $G \simeq H$ and $G$ is abelian, then $H$ is abelian", which is equivalent to saying "If $H$ is not abelian then either $G$ is not abelian or $G \not\simeq H$". Since we have assumed that $G$ is abelian, we know that $G \not\simeq H$, and we are done.