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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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    I have figured out that if a is in the center of G, then $ø_a(x) = axa^{-1} = aa^{-1}x = x$, so to find the example I just have to pick an element such that some power of that element is in the center. Obviously $ø_a^{|a|}$ is the identity transformation but how can I know if this is the smallest such power?2012-10-30
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    @ChadRussell: That is a good idea for finding the example. I've just updated before I saw your comment with another way. As for your question, $|a|$ is *not* necessarily the smallest such power. If $\phi_a^n$ is the identity transformation, what does that tell you about the relationship between $n$ and the order of $\phi_a$ (second bullet)?2012-10-30
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    @ChadRussell: Your comment brings to mind the [quaternion group](http://en.wikipedia.org/wiki/Quaternion_group).2012-10-30
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    If $ø_a^n$ is the identity, then $|ø_a| | n$. I still do not know how to relate this n value to the order of a, but you have gotten me farter than I ever would have gotten on my own so I'll mark it solved. Thank you!2012-10-30
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    @Chad: That is a true general statement; it holds for all $n$ such that $\phi_a^n$ is the identity. In particular, it holds in the special case where $n = |a|$, based on your observations.2012-10-30
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    aha! the order of an element divides the order of a group, forgot about that one.2012-10-30
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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