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The question is. Let $G$ be a group and let $a,b\in G$ be two fixed elements which happen to commute $(ab=ba)$. Prove that $H=\{x\in G \mid axb=bxa\}$ is a subgroup of $G$.

The book goes over the proof: $axb=bxa$ implies that $x(ba^{-1})=(a^{-1}b)x$. And since $ab=ba$, it follows that $ba^{-1}=a^{-1}b$. So, $H$ is a subgroup.

I was wondering if someone could walk me through a more formal proof for a subgroup with this (i.e showing it commutes, holds for inverses, etc.).

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    $\begingroup$ Just because this doesn't do the usual tests directly does not mean it's not a "formal" proof. This is just as rigorous, and arguably better if there's no obvious way to check the closure properties without symbolic manipulations that mimic this proof anyway. (Also, how is "showing it commutes" part of showing a subset is a subgroup, and what does "it" in this phrase refer to anyway?) $\endgroup$ – anon Dec 16 '13 at 17:18
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    $\begingroup$ (Also, as Ian notes, this proof is not quite finished; one should for example remark that $H$ is the set of fixed points of a particular conjugation automorphism, and either argue or invoke the fact that fixed points of an automorphism pass all the subgroup test.) $\endgroup$ – anon Dec 16 '13 at 17:24
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It doesn't look like what you posted is any kind of a proof. Here's what you'd like:

Claim: $e\in H$. Well since $ab=ba$, $aeb=ab=ba=bea$, so that's fine.

Claim: for $x,y\in H$, $xy\in H$. Well $axyb=(axb)b^{-1}a^{-1}(ayb)=(bxa)b^{-1}a^{-1}(bya)$. Now since $a,b$ commute, their inverses do too. So we hae $axyb=(bxa)a^{-1}b^{-1}(bya)=bxya$, as required.

Other strategy: $H$ is precisely the centraliser of the element $b^{-1}a$. Centralisers are subgroups. Therefore $H$ is a subgroup.

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    $\begingroup$ Possibly the book proof assumes that the reader remembers that centralizers are subgroups? $\endgroup$ – Henning Makholm Dec 16 '13 at 17:26
  • $\begingroup$ Yes instead of showing the axioms once again, one just has to remember that one has already shown that centralizers are subgroups (probably this has been done in the referenced text). $\endgroup$ – Martin Brandenburg Dec 16 '13 at 17:29

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