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Let G be an abelian group. Prove that

$$G^{(n)} = \{g \in G | g^n = 1_G \}$$

is a subgroup of G.

How do I go about doing this?

I understand that $G^{(n)}$ is basically the set of all elements whose order divides n. So would I simply need to show that the group axioms hold for $G^{(n)}$ and this would show that it is a subgroup for G?

I already have the identity element at n = 0. Also, the inverse would be any $n < 0$. So now I just have to prove associativity right? How do I go about doing this?

Is this when $g^{(n)^{(m)}} = g^{(m)^{(n)}}$? How would I prove that?

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Just a note: That notation is not completely universal, in that $G^{(n)}$ is often used to denote the $n$'th derived (ie, commutator) subgroup. – Tobias Kildetoft Oct 17 '12 at 11:11
@Tobias In the abelian case, at least according to my sources, this subgroup is usually written $nG$, in the non-abelian case it's more common to use $G^n$, without parentheses. – Arthur Oct 17 '12 at 11:19
@Arthur What subgroup? The subgroup $nG$ is neither of the subgroups discussed here. It is the subgroup consisting of $n$'th powers (written additively for an abelian group) of elements from $G$. For non-abelian $G$, $G^n$ might refer either to the subset of such powers or to the subgroup generated by them. – Tobias Kildetoft Oct 17 '12 at 11:25
0% accepted rate means no good here. ;-) – Nancy Rutkowskie Oct 17 '12 at 11:27
@Tobias Yes, you're right. I don't know what I was thinking. – Arthur Oct 17 '12 at 12:02
up vote 5 down vote accepted

To show that $H$ is a subgroup of $G$, you only need to show that

  1. $1_G \in H$,
  2. If $g, h \in H$, then $gh \in H$, and
  3. If $g \in H$, then $g^{-1} \in H$.

Some people prefer to combine 2 and 3 together and just prove

  • If $g, h \in H$, then $gh^{-1} \in H$.

You can use whichever way you find easier. I'll prove all 3 things for the case $H = G^{(n)}$.

  1. $1_G \in G^{(n)}$ because $1_G^n = 1_G$.
  2. If $g, h \in G^{(n)}$, then $(gh)^n = g^n h^n = 1_G 1_G = 1_G$, so $gh \in G^{(n)}$.
  3. If $g \in G^{(n)}$, then $(g^{-1})^n = (g^n)^{-1} = 1_G^{-1} = 1_G$, so $g^{-1} \in G^{(n)}$.

Therefore, $G^{(n)}$ is a subgroup of $G$. Note that we used the fact that $G$ is abelian in the proof of 2.

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$G^{(n)}$ is a subset of $G$, that is, its elements are elements of $G$ and so associativity is automatically satisfied. The identity $1_G$ is in $G^{(n)}$ because $1_G^n=1_G$, and if $g \in G^{(n)}$, that is, $g^n = 1_G$, then $(g^{-1})^n = (g^n)^{-1} = 1_G^{-1} = 1_G$, so that $g^{-1}$ is also in $G^{(n)}$. Finally, we need to show $G^{(n)}$ is closed under composition. Now, if $g,h \in G^{(n)}$, then $(gh)^n = g^nh^n = 1_G$, where the first equality comes from the assumption that $G$ is abelian, and the second comes from the assumption $g,h \in G^{(n)}$. Thus, all requirements for $G^{(n)}$ to be a subgroup of $G$ are satisfied.

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What about the product of two elements of $G^{(n)}$? – Alexander Thumm Oct 17 '12 at 11:09
Oh, sorry I forgot, I guess that's where the abelian assumption comes in. If $g, h \in G^{(n)}$, then $(gh)^n = g^nh^n = 1_G$, where the first equality requires the abelian assumption, and so $gh \in G^{(n)}$. – HJSprime Oct 17 '12 at 11:14
I think adding an "edit" to this is the most convenient thing to to as it is missing the main ingredient: showing closure of operation, which requires abelian. – DonAntonio Oct 17 '12 at 11:17
@DonAntonio Okay sure, I've edited it now. Sorry about that, I'm new here. – HJSprime Oct 17 '12 at 11:20
Don't be sorry, we've all done things like that. This time your answer is complete and thus +1 – DonAntonio Oct 17 '12 at 11:24

Another way to go:

$G$ is an abelian group. What can we say about the map $f: G \rightarrow G$ defined by $f(g)=g^n$? What would be the kernel of this map?

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