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Prove that in every group $G$, $\forall a,b \in G$, $O(a)=O(bab^{-1})$.

NOTE: The $O()$ notation is used to represent the order of a group (i.e if $O(a)=3$ then $a^3=e$ (the identity element) ). I am highlighting this because I'm not sure if this notation is widely used.

Now so far I've only managed to prove that $m|n$ from

$(bab^{-1})^n=bab^{-1}bab^{-1}\ldots bab^{-1}=baeae...eab^{-1}=ba^nb^{-1}=bb^{-1}=e \implies m|n$.

On the other hand, from

$a^m=a\cdot a\cdot a\cdot...\cdot a$

And this is where I'm stuck. How can I rewrite $a^m$ to include $bab^{-1}$?

EDIT: I've never heard of homomorphism nor do I know how to make use of it.

downmath
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2 Answers2

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If $o(bab^{-1})=m$ and $o(a)=n$ then,

$(bab^{-1})^m=e\implies ba^mb^{-1}=e\implies a^m=b^{-1}eb=e\implies n|m.$

Koro
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Another version of what @Koro noted is to use the following theorem:

Theorem: Let $f:G_1\to G_2$ be a group homomorphism which is also a 1-1 map. So $|g|=|f(g)|$ where $g\in (G_1,e_1)$.

Now, you can take $f=f_g:G\to G$ where $f_g(x)=g^{-1}xg$.

Mikasa
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