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$a,b$ are group elements. If $a$ commutes with $b$, does $a^{-1}$ commute with $b$?

And how can I prove this?

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21  
$a^{-1}b = a^{-1}baa^{-1}$. –  Daniel Fischer Aug 11 '13 at 16:39
    
math.stackexchange.com/questions/89079/… is more or less a duplicate, or at least of interest –  Jack Schmidt Aug 11 '13 at 18:06

4 Answers 4

up vote 16 down vote accepted

$$\begin{align*} ab=ba &\Leftrightarrow aba^{-1}=baa^{-1}\\ &\Leftrightarrow aba^{-1}=b\\ &\Leftrightarrow a^{-1}aba^{-1}=a^{-1}b\\ &\Leftrightarrow ba^{-1}=a^{-1}b \end{align*}$$

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1  
Dat last line typo 8) –  Patrick Da Silva Aug 11 '13 at 16:42
    
Gottit${}{}{}{}$ –  user1729 Aug 11 '13 at 16:44
    
@PatrickDaSilva LAWL. –  Pedro Tamaroff Aug 11 '13 at 18:05

Note that if $$ab=ba$$ then multiply previous relation by $a^{-1}$ from the left to get: $$b=e_Gb=(a^{-1}a)b=a^{-1}(ab)=a^{-1}(ba)=a^{-1}ba$$ Now , again multiply previous relation by $a^{-1}$ from the right to get: $$ba^{-1}=(a^{-1}ba)a^{-1}=a^{-1}b(aa^{-1})=a^{-1}be_G=a^{-1}b$$

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It deserves +1. –  Sami Ben Romdhane Aug 11 '13 at 16:48
    
(+1) nice answer. –  Mhenni Benghorbal Aug 11 '13 at 18:39
    
Looks like a "nice answer"! +1 –  amWhy Aug 11 '13 at 23:57
    
@MahdiKhosravi: $\ddot\smile$ –  Babak S. Aug 15 '13 at 11:53

More generally, for subsets $U\subseteq G$, the normalizer of $U$ is defined as $$ N_G(U):=\{\,g\in G\mid gU=Ug\,\}$$ and is a subgroup of $G$ because

  • $eU =U=Ue$ trivially
  • $gU=Ug $ and $hU=Uh$ implies $(gh)U=g(hU)=(gU)h=(Ug)h=U(gh)$
  • $gU=Ug$ implies $g^{-1}U=g^{-1}Ugg^{-1}=g^{-1}gUg^{-1}=Ug^{-1}$

that is $N_G(U)$ is nonempty and closed under multiplication and taking inverses.

You are concerned with the special case $U=\{b\}$ and the claim $a\in N_G(U)\implies a^{-1}N_G(U)$.

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Thanks for the enlightening generalization! –  parsec Aug 11 '13 at 16:59

@parsec Now you have seen all the answers you probably can prove that any power $a^n$ commutes with $b$, in other words, the group generated by your $a$ and $b$, $<a,b>$, is abelian.

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Thank you for the reminder! I've proved the $a^n$ case before but I didn't realize the connection. –  parsec Aug 13 '13 at 11:29

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