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Here is what I did:

Suppose R $\leq$ $D_n$ is the subgroup of rotation. Suppose $x \in fRf^{-1}$. Then $x = frf^{-1}$ for some r and for any r. Therefore, $x = r'$ a rotation and hence $x \in R$. therefore by definition R is normal in $D_n$

Is this correct and complete verification?

Thanks, any other suggestions??

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Any subgroup of index 2 of any group is normal. –  Gerry Myerson Oct 5 '12 at 7:20
    
Thanks Gerry, this is faster. –  Chasky Oct 5 '12 at 7:21
    
but the argument above is correct? –  Chasky Oct 5 '12 at 7:39
    
I don't quite understand your statement "Then $x = frf^{-1}$ for some $r$ and for any $r$". There are certainly more elegant ways of doing it, but my personal approach would be to show that $x$ is a rotation by using the fact that every $f \in D_n$ is generated by the elements $r,s$ satisfying $s^2 = 1$ and $srs = r^{-1}$. –  Christopher A. Wong Oct 5 '12 at 7:58

1 Answer 1

up vote 2 down vote accepted

Well, it is correct. But you should emphasize and expand a bit more the essence of the thought, and this is:

Therefore $x=frf^{-1}$ is a rotation.

So.. Why is it a rotation? That's the point.

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