Elements of odd-order abelian groups are squares.

  • Thread starter Kreizhn
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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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Answers and Replies

  • #2
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Hi Kreizhn! :smile:

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...
 
  • #3
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A little pedantry:

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

This should be |<g>| | |G|.
 
  • #4
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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 [itex] g \mapsto g^k [/itex] is surjective.

I haven't finished it yet, but we need the group to be abelian for [itex] g \mapsto g^k [/itex] 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:
 
  • #5
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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
[tex] G = \bigcup_{g \in G} \langle g \rangle [/tex]?
If this is the case, then we again need only show that [itex] g \mapsto g^k [/itex] is surjective on the cyclic subgroups.
 
  • #6
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Hmm, the books seems to contain a lot of typos :smile:
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...
 
  • #7
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Wait a tick. I don't think we need abelian in the second question either.



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

Yes, that's what I had in mind :smile:
 
  • #8
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This is what I have.

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

[tex] \phi(g^{ar}) = g^{akr} = g^{r-bm|g|r} = g^r (g^{|g|})^{bmr} = g^r [/tex]

and so the function is surjective on cyclic subgroups.

Here's the only thing that jumps in my head. Hopefully it's trivial. Since [itex] \phi\bigg|_{\langle g \rangle} [/itex] 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.
 
  • #9
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Seems ok! :smile:
 

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