# Order of conjugate of an element given the order of its conjugate

Let $G$ is a group and $a, b \in G$. If $a$ has order $6$, then the order of $bab^{-1}$ is...

How to find this answer? Sorry for my bad question, but I need this for my study.

• Hint: conjugation is an automorphism – dani_s Jan 10 '14 at 14:27
• @dani_s: that would be too much... it is a basic question and you are proposing to look at the automorphism.. Not a good idea i guess.. – user87543 Jan 10 '14 at 15:32

$$|bab^{-1}|=k\to (bab^{-1})^k=e_G$$ and $k$ is the least positive integer. But $e_G=(bab^{-1})^k=ba^kb^{-1}$ so $a^k=e_G$ so $6\le k$. Obviously, $k\le 6$ (Why?) so $k=6$.

Two good pieces of advice are already out here that prove the problem directly, but I'd like to decompose and remix them a little.

For a group $G$ and any $g\in G$, the map $x\mapsto gxg^{-1}$ is actually a group automorphism (self-isomorphism). This is a good exercise to prove if you haven't already proven it.

Intuitively, given an isomorphism $\phi$, $\phi(G)$ looks just like $G$, and $\phi(g)$ has the same group theoretic properties as $g$. (This includes order.) This motivates you to show that $g^n=1$ iff $\phi(g)^n=1$, and so (for minimal choice of $n$) they share the same order.

Here's a slightly more general statement for $\phi$'s that aren't necessarily isomorphisms. Let $\phi:G\to H$ be a group homomorphism of finite groups. Then for each $g\in G$, the order of $\phi(g)$ divides the order of $g$. (Try to prove this!)

If $\phi$ is an isomorphism, then so is $\phi^{-1}$, and so the order of $\phi(g)$ divides the order of $g$, and the order of $\phi^{-1}(\phi(g))=g$ divides the order of $\phi(g)$, and thus they're equal.

• @Andreas It seems this question (and variants) are destined to be prototypical examples of an abstract duplicate (e.g. recall the recent question). In fact, even the comments are becoming duplicate! – Bill Dubuque Jan 12 '14 at 17:56
• @BillDubuque, an optimistic view of the fact that the comments are becoming duplicates is that we are reaching a consensus on a canonical form for answers and comments ;-) – Andreas Caranti Jan 12 '14 at 18:08

Note that $(bab^{-1})^2 = bab^{-1}bab^{-1} = ba^2b^{-1}$. Similarly $(bab^{-1})^n = ba^nb^{-1}$ for any $n$. When will $ba^nb^{-1} = 1$ using the information about $a$? Then you just have to check to see that $ba^mb^{-1} \not = 1$ for any $1 \leq m < n$.

• I still don't get it. – Yagami Jan 10 '14 at 14:58

In general, let $o(a)=n$ and $o(bab^{-1})=k$, then $(bab^{-1})^k=ba^kb^{-1}=e$, by Cancellation Law in group, we can get $a^k=e$, since $o(a)=n$, then $k \geq n$ (in fact we can get $n|k$, but in this proof $k \geq n$ is enough). Easy to see that if $k=n$ then $(bab^{-1})^n=ba^nb^{-1}=beb^{-1}=e$, hence $k=n$.

CONCLUSION: $o(a)=o(bab^{-1})$.