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Suppose that G is a finite, nonabelian group with odd order. Show s is surjective, and hence bijective.

I have been told to look at the effects of the squaring map, $s\colon G\to G$, defined by $s(g)=g^2$ on the elements of cyclic groups $\langle g\rangle$ of $G$.

I'm stumped. Could anyone give me a nudge in the right direction or (being hopeful) a full solution?

Thanks a lot.

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The original version of this question on MO had a good hint. Have you tried using that? –  Qiaochu Yuan Aug 16 '11 at 20:51
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The title contains a lot of information that is absent from the body (e.g., that $G$ is nonabelian of odd order). Please make the body of the post self-contained. The title is an indexing feature, much like the spine of a book. I don't know many books that ask you to read the spine in order to be able to understand what the author is saying. –  Arturo Magidin Aug 16 '11 at 20:53
    
Try this by hand on cyclic groups of order 3,5,7,... until you see the light. –  Jyrki Lahtonen Aug 16 '11 at 20:54
    
Hint: The order of $g$ divides the order of $G$, so it must be odd. If the order of $g$ is $n$, then the order of $g^k$ is $n/\gcd(n,k)$. –  Arturo Magidin Aug 16 '11 at 20:54
    
Okay thanks guys, I'll have a think and post back later if I figure it out (I should be able to) Also, @Qiaochu, I could not see any hints on MO. I just saw 2 comments telling me to come here –  Chris Aug 16 '11 at 21:25

1 Answer 1

up vote 1 down vote accepted

On each cyclic subgroup $C$ of $G$, the $s$ map is a homomorphism. Prove that $s$ is injective by considering $\ker s$ in $C$. It is here that you use that $G$ of odd order.

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Say x = ker(s), does s(x) = e? So x^2=e which can be of order 1 or 2, but as we know the elements can only be odd order? I used this for a previous question, the difference being G was abelian. My problem is I'm not sure how this relates to the cyclic groups. I can see the effect when I apply the map s to elements of odd cyclic groups but not sure how to apply that to my answer... –  Chris Aug 16 '11 at 21:54
    
@Chris, you're on the right track. The point is that $s$ is injective in $C$ and so is surjective in $C$. Hence $s$ is surjective in $G$ because every element of $G$ is in a suitable $C$ (the one generated by itself). –  lhf Aug 16 '11 at 21:58

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