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Let $a$ belong to a group $G$ and let $|a|$ be finite. Let $ø_a(x) = axa^{-1}$ for elements $x$ in $G$. Show that the order of $ø_a$ divides the order of $a$. Exhibit an element from a group for which $1 < |ø_a| < |a|$.

Since this is technically homework I'm not expecting a complete proof, but an idea of where to start would be greatly appreciated.

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  • What is $\phi_a^{|a|}$?

  • If $g$ is an element of a group and $g^n$ is the identity element of the group, then what does this tell you about the relationship between $n$ and $|g|$?

I forgot about the second part:

If you've seen direct products of groups, here's an approach. First, can you think of a group $H$ with an element $b$ of order $3$, but for which $\phi_b$ is the identity automorphism? Next, can you think of a group $K$ with an element $c$ of order $2$ for which $\phi_c$ also has order $2$? If so, then these can be pieced together on $H\times K$, with $a=(b,c)$.

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    @Chad: That is related but not quite the point. It is just the fact that if $g^n$ is the identity, then the order of $g$ divides $n$. $n$ need not be the order of any group.2012-10-30