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## Homework Statement

Let G be a finite abelian group, and assume that |G| is odd. Show that every element of G is a square.

## The Attempt at a Solution

So we want to show that [itex] \forall g \in G, \exists h \in G, g = h^2 [/itex]. Let [itex] g \in G [/itex] be arbitary, and consider the subgroup generated by g, denoted <g>. By Lagrange's theorem, [itex] |<g>| \big| |G| [/itex] and so the order of g (which I will denote by |g|) must also divide |G|. Since |G| is odd, this means that |g| is also odd. In particular, choose [itex] n \in \mathbb N [/itex] such that |g| = 2n+1. Then

[tex] g^{2n+1} = g^{2n} g = (g^n)^2 g = e. [/tex]

Define [itex] h=g^{-n} [/itex] so that this becomes

[tex] (h^{-1})^2 g = h^{-2} g = e [/tex]

so that the result follows.

Now I can't see where I've used the fact that G is abelian, which makes me think I've done something wrong. Can anyone point out the error, if there is one?

Edit: Fixed mistake, [itex] h = g^{-n} [/itex].

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