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Let $G$ be a group and let $f:G\to G$ be an isomorphism. Let $H = \{ a \in G \; | \; f(a)=a^{-1}\}$ Prove that $H$ is a subgroup of $G$ if and only if $G$ is abelian.

My attempt: I already know that $H$ is a subset of $G$ since $a$ is in $G$. And that $H$ is $a$ group if and only if for every $a$ and $b$ in $H$, $ab^-1$ is in $H$. That is, if $f(a)=a^{-1}$ and $f(b)=b^{-1}$ then $f(ab^-1)=(ab^{-1})^{-1}=(b^{-1})a$.

But I don't know how to derive this nor the converse based on the abelian property of $G$ that for every $a$ and $b$ in $G$, $ab=ba$.

Please help some. Thanks!

Miguelgondu
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Jeffrey
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2 Answers2

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Suppose $G$ is abelian. Let $a, b \in H$. Then $$f(ab^{-1}) = f(a)f(b)^{-1} = a^{-1}(b^{-1})^{-1} = (b^{-1}a)^{-1} = (ab^{-1})^{-1}$$ which implies $ab^{-1}\in H$. Since $a$ and $b$ were arbitrary, $H$ is a subgroup of $G$.

The converse need not hold. If $G$ is the nonabelian group $\Bbb Z/7\Bbb Z \rtimes \Bbb Z/3\Bbb Z$, and $f: G \to G$ is given by $f(g) = g$, then $H$ consists of the identity together with all the elements of $G$ of order $2$. Since $G$ has no elements of order $2$, $H$ consists only of the identity. So $H$ is a subgroup of $G$.

kobe
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  • But such elements are just in $H$, shouldn't we start from considering $a,b \in G$ and not just $a,b\in H$ to prove that $G$ is abelian? – Miguelgondu Feb 04 '15 at 15:55
  • @Miguelgondu I forgot to add that $G$ is the homomoprhic image of $H$. Sorry about that. – kobe Feb 04 '15 at 16:14
  • How come?, is there an epimorphism from $G$ onto $H$? – Miguelgondu Feb 04 '15 at 16:28
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    @Miguelgondu I meant the other way around, but it doesn't matter (since $f$ is an isomorphism). I gave an example of how the converse doesn't hold. – kobe Feb 04 '15 at 17:54
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I don't think the statement is true. Consider any non-abelian group $G$ whose order does not divide $2$ and the isomorphism given by the identity map $f(a)=a$, then $H=\{id\}$ is a subgroup...

I didn't misunderstand the question or screw up the example did I?

Qidi
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