# Elements of odd-order abelian groups are squares.

Kreizhn

## 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 $\forall g \in G, \exists h \in G, g = h^2$. Let $g \in G$ be arbitary, and consider the subgroup generated by g, denoted <g>. By Lagrange's theorem, $|<g>| \big| |G|$ 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 $n \in \mathbb N$ such that |g| = 2n+1. Then
$$g^{2n+1} = g^{2n} g = (g^n)^2 g = e.$$
Define $h=g^{-n}$ so that this becomes
$$(h^{-1})^2 g = h^{-2} g = e$$
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, $h = g^{-n}$.

Last edited:

## Answers and Replies

Staff Emeritus
Homework Helper
Hi Kreizhn! Your proof looks good. In fact, you proved the slightly more general theorem

if g is an element of group G and if g has odd order, then g is a square.​

I don't quite see why they assume G to be abelian, though. In fact, the theorem for abelian groups implies the theorem for general groups. Indeed, take g in a general group G with odd order, then <g> is abelian with odd order and the theorem applies...

Staff Emeritus
Homework Helper
A little pedantry:

Kreizhn;3345909 By Lagrange's theorem said:
<g> \big| |G| [/itex]

This should be |<g>| | |G|.

Kreizhn
A little pedantry:
This should be |<g>| | |G|.

Thanks, I fixed it. That's what I thought. I wonder if maybe the author didn't just get overzealous or make a funny mistake. The next question in the book I'm working through is a generalization of that question, and I think abelianism might be needed there.

Let G be a finite (abelian) group with order |G| = n, and take k any integer coprime to n. Show that the mapping $g \mapsto g^k$ is surjective.

I haven't finished it yet, but we need the group to be abelian for $g \mapsto g^k$ to be a group homomorphism and I figure that's going to be important.

In an interesting note, he actually puts abelian in the first question, but NOT in the second. I have his errata in front of me, which is how I know (abelian) is supposed to show up for the generalization. Maybe he just put abelian in the wrong spot :rofl:

Kreizhn
Wait a tick. I don't think we need abelian in the second question either.

I don't quite see why they assume G to be abelian, though. In fact, the theorem for abelian groups implies the theorem for general groups. Indeed, take g in a general group G with odd order, then <g> is abelian with odd order and the theorem applies...

Is the point here that we can write
$$G = \bigcup_{g \in G} \langle g \rangle$$?
If this is the case, then we again need only show that $g \mapsto g^k$ is surjective on the cyclic subgroups.

Staff Emeritus
Homework Helper
Hmm, the books seems to contain a lot of typos I'm not even sure if abelianness is needed for the second assertion. We need it for being a homomorphism, but perhaps not for surjectivity...

Staff Emeritus
Homework Helper
Wait a tick. I don't think we need abelian in the second question either.

Is the point here that we can write
$$G = \bigcup_{g \in G} \langle g \rangle$$?
If this is the case, then we again need only show that $g \mapsto g^k$ is surjective on the cyclic subgroups.

Yes, that's what I had in mind Kreizhn
This is what I have.

Since $\gcd(k,n) = 1$ it is necessary that $\exists a,b \in \mathbb Z$ such that ak + bn = 1. Let $g \in G$ be an arbitary element, and consider the cyclic subgroup $\langle g \rangle$. By Lagrange's theorem, $\exists m \in \mathbb Z$ such that n = |g| m and so
$$ak + bm|g| = 1.$$
Now we claim that $\phi: G \to G, g \mapsto g^k$ is surjective. Since the cyclic subgroup is abelian, $\phi\bigg|_{\langle g \rangle}$ is a homomorphism. We want to show that $\phi\bigg|_{\langle g \rangle}$ is surjective on $\langle g \rangle$. Indeed, let $g^r \in \langle g \rangle$, then

$$\phi(g^{ar}) = g^{akr} = g^{r-bm|g|r} = g^r (g^{|g|})^{bmr} = g^r$$

and so the function is surjective on cyclic subgroups.

Here's the only thing that jumps in my head. Hopefully it's trivial. Since $\phi\bigg|_{\langle g \rangle}$ is surjective on all cyclic subgroups, it's surjective as a set mapping on the entire group G. However, it needn't be a homomorphism. Luckily, we were never asked that it be a homomorphism.

Staff Emeritus
Seems ok! 