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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?

- Non-isomorphic Group Structures on a Topological Group
- Can finite non-isomorphic groups of the same order have isomorphic endomorphism monoids?
- Finite groups with exactly one maximal subgroup
- Cyclic group generators
- Wielandt's proof of Sylow's theorem.
- Why is the quotient map $SL_n(\mathbb{Z})$ to $SL_n(\mathbb{Z}/p\mathbb Z)$ is surjective?

Thanks a lot.

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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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