Let G be an abelian group. Show $\{ x \in G | x^n=e$ for some $n\in \mathbb{Z} \}$ is a subgroup of G.

$\textbf{Proof:}$ Assume G is an abelian group. So G has an identity element, each element has a unique inverse, G has closure, G is associative, and if $a,b \in G$ then $ab=ba$. Let $$A=\{ x \in G | x^n=e \, \space\text{ for some } \space n\in \mathbb{Z} \}$$ We need to show that A is a subgroup of G. (i.e. A has an identity element, each element has a unique inverse, and G has closure.

We don't need to show that A has an identity element because the group description shows us that $x^n$ is the identity element since $x^n=e$. We also know because G has closure a subset of G, A, will also have closure.

So what is left to show is that A has an inverse. By the definition of an inverse, \begin{equation*} \begin{aligned} x^nx^{-1} &=e=x^n \iff x^{-1}=1\\ x^{-1}x^n &=e=x^n \iff x^{-1}=1\\ \end{aligned} \end{equation*} Hence A has an inverse. So A is a subgroup of G.

Is this correct?

  • $\begingroup$ You could clarify the "We don't need to show that $A$ has an identity..." by noting that $e\in G,e^1\in G\implies e\in A$... The group description did not tell you that $x^n\in A$, only that $x\in A$. $\endgroup$ – abiessu Oct 9 '14 at 15:42
  • $\begingroup$ This is not correct - subsets of $G$ don't automatically have closure. Consider the group $G=(\mathbb{Z}, +)$; then $A=\{1, 3, 5\}$ is a subset of $G$, but it certainly isn't closed under $+$; $1+3=4\not\in A$... $\endgroup$ – Steven Stadnicki Oct 9 '14 at 15:42
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    $\begingroup$ Note that $e^n=e$ for all $n$ $\endgroup$ – AsdrubalBeltran Oct 9 '14 at 15:43
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    $\begingroup$ Your proof of the identity isn't great, you need to show that $e$ satisfies the defining property of the group. Also, you haven't shown that $A$ is closed under multiplication. So far your argument hasn't appealed to the fact that $G$ is abelian, which should be a hint that you aren't done. $\endgroup$ – James Oct 9 '14 at 15:43
  • $\begingroup$ Also $(x^{-1})^n=(x^n)^{-1}=e^{-1}=e$ $\endgroup$ – AsdrubalBeltran Oct 9 '14 at 15:47

A subset of a group doesn't necessarily have closure. To show closure, you need to select two arbitrary elements of $A$, say $x, y \in A$, and show that $xy \in A$.

Let $x, y \in A$. Then $x^n = e, y^n \in e$, where $n\in \mathbb Z$. Then $x^ny^n = ee = e = (xy)^n$, because $G$ is abelian. Hence, $xy\in A$. Thus, $A$ is closed under the group operation of $G$ (and hence under the group operation of $A$).

To show that $A$ is closed under inverses, you need to show that for an arbitrary element $x\in A$, that $x^{-1} \in A$ as well. (We know that there exists an inverse $x^{-1} \in G$, we just need to show that it is also in $A$ if $A$ is a subgroup of $G$.

Let $x \in A$. So, $x^n = e$. Hence $(x^{-1})^n = (x^n)^{-1} = e^{-1} = e$. Indeed, $x^{-1}\in A$.

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    $\begingroup$ $x$ and $y$ don't have to be annihilated by the same power $n$... $\endgroup$ – Steven Stadnicki Oct 9 '14 at 15:55
  • $\begingroup$ But $n = \operatorname{lcm} \{ n_x, n_y \}$ does work. $\endgroup$ – Sammy Black Oct 10 '14 at 2:08

A hint towards showing closure:

Suppose $a\in A$; then you know some power of $a$ is the identity, let's say $a^j=e$. Similarly, if $b\in A$, then there's some power of $b$ that's the identity, say $b^k=e$.

Now, since any power of the identity is the identity (why?), then for any $m$ it's the case that (for instance) $(a^j)^m=a^{jm}=e$.

What's more, since $G$ is abelian then we know that $(ab)^n = a^nb^n$. Can you find some value of $n$ that you know will be guaranteed to make both terms on the right the identity? (And can you see why this shows closure?)


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