If in a group $G$, we have $a^5 = e$ and $aba^{-1} = b^2$ for some $a$, $b$ in $G$, then what is the order of $b$? Here $e$ denotes the identity element in $G$.


It is not hard to see that $a^nba^{-n}=b^{2^n}$. Thus, $b^{31}=1$. Note that $31$ is prime and so $o(b)=31$ or $b=1$.

  • $\begingroup$ Babak Miraftab, there are bound to be other solutions, but what is important is the fact that this solution is so simple and uses only the elementary facts about groups. $\endgroup$ – Saaqib Mahmood Apr 2 '14 at 13:20
  • $\begingroup$ Can we find a concrete example of a group having two elements satisfying these conditions? $\endgroup$ – Saaqib Mahmood Apr 2 '14 at 13:24
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    $\begingroup$ @user127001 Note that $a^2ba^{-2}=a^1b^2a^{-1}=(a^1ba^{-1})^2$. $\endgroup$ – Babak Miraftab Apr 2 '14 at 13:51
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    $\begingroup$ @user127001 Put $n=5$ in the equation. $\endgroup$ – Braindead Apr 2 '14 at 13:54
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    $\begingroup$ @BabakMiraftab: chera man ino nadide budam. Good to go for 10 :) $\endgroup$ – Mikasa Aug 13 '14 at 11:24

Here is a more general result.

If $G$ is a group and $a$,$b$ $\in$ $G$ with $bab^{-1} = a^r$ for some $r$ $\in$ $\mathbb{N}$, then $b^jab^{-j} = a^{r^j}$ for all $j$ $\in$ $\mathbb{N}$

Proof : We use induction. The base case is trivial. Let the hypothesis be true for some natural number $k$. We have :

$b^{j+1}ab^{-(j+1)} = b(b^jab^{-j})b^{-1} = ba^{r^j}b^{-1} = (bab^{-1})^{r^j} = (a^r)^{r^j} = a^{r.r^j} = a^{r^{j+1}}$


Now you can proceed as Babak did. This is taken from an exercise in Hungerford's Algebra.


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