# Elements of odd-order abelian groups are squares.

• Kreizhn
In summary, we want to show that if G is a finite abelian group with odd order, then every element of G is a square. Using Lagrange's theorem and the fact that |G| is odd, we can show that the order of any element must also be odd. This means that we can choose a certain power of the element to be the identity, and by defining a new element, we can show that the original element is a square. Abelianness is not necessary for this proof, as shown by the fact that the author mistakenly included it in the first question.
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:
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...

A little pedantry:

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

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

micromass said:
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

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

micromass said:
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.

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...

Kreizhn said:
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

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.

Seems ok!

## 1. What does it mean for an element of an odd-order abelian group to be a square?

Being a square in an abelian group means that the element can be expressed as the product of two copies of itself. In other words, it has a square root within the group.

## 2. Why are elements of odd-order abelian groups specifically mentioned in this statement?

This statement specifically refers to odd-order abelian groups because in even-order groups, not all elements have square roots. This is due to the fact that for an element to have a square root, its order must be a multiple of 2.

## 3. Can you provide an example of an odd-order abelian group where this statement holds true?

One example of an odd-order abelian group where this statement holds true is the cyclic group of order 5, denoted by Z5. In this group, the only elements are 1, 2, 3, 4, and 5, and every element can be expressed as a square (e.g. 3 = 2 * 2).

## 4. How does this statement relate to the concept of roots in mathematics?

This statement relates to the concept of roots in mathematics by showing that within an odd-order abelian group, every element has a square root, just as every number has a square root in mathematics.

## 5. Why is this statement important in the study of abelian groups?

This statement is important in the study of abelian groups because it helps to characterize and understand the structure of these groups. It also has applications in other areas of mathematics, such as number theory and algebraic geometry.

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