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?

  • 1
    $\begingroup$ Do you know what "isomorphic" means? $\endgroup$ – Chris Eagle Dec 2 '12 at 20:30
  • $\begingroup$ Yes, an isomorphism is a bijective homomorphism. $\endgroup$ – Onur Dec 2 '12 at 20:31

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.

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Let $N$ be the non-abelian group and $A$ be the abelian group.

Take $(a, b)\in N^2$ that do not commute, ie $ab \not= ba$.

Let $f$ be your bijection from $N$ to $A$.

$f(ab)=f(a)f(b)=f(b)f(a)=f(ba)$ because $f(a)$ and $f(b)$ are in $A$ which is abelian.

So since $f$ is a bijection, $ab = ba$, absurd.

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Suppose $\,f:A\to G\,$ is a group isomorphism, with $\,A\,$ abelian, $\,G\,$ non-

abelian, then

$$\exists\,x,y\in G\,\,s.t.\,\,xy\neq yx$$

But since $\,f\,$ is bijective there exist $\,a,b\in A\,$ s.t.

$$f(a)=x\,\,,\,\,f(b)=y\Longrightarrow xy=f(a)f(b)=f(ab)=f(ba)=f(b)f(a)=yx ...$$

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