6
$\begingroup$

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 $G$.

$\endgroup$
16
$\begingroup$

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 $b=1$.

$\endgroup$
8
  • $\begingroup$ 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
  • 1
    $\begingroup$ @user127001 Note that $a^2ba^{-2}=a^1b^2a^{-1}=(a^1ba^{-1})^2$. $\endgroup$ – Babak Miraftab Apr 2 '14 at 13:51
  • 2
    $\begingroup$ @user127001 Put $n=5$ in the equation. $\endgroup$ – Braindead Apr 2 '14 at 13:54
  • 1
    $\begingroup$ @BabakMiraftab: chera man ino nadide budam. Good to go for 10 :) $\endgroup$ – Mikasa Aug 13 '14 at 11:24
3
$\begingroup$

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 :

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

QED.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.