Show that H is a subgroup of G

Question

Let n be a positive integer and let G be an Abelian group. Define

$H := \{x ∈ G : |x| \ \text{divides} \ n\}$

where x denotes the order of |x|. Show that H is a subgroup of G.

I know I need to prove that H is non-empty, contains the identity, e, and that it contains $ab^{-1}$. I'm just not sure how to prove this, thank you for any help!

Answer 1

Let me try. We have $e \in H$, because $|e| = 1$ divides $n$.

Let $x$, $y$ in $H$, then $|x|$, $|y|$ divides $n$. We need to prove that $|xy^{-1}|$ divides $n$. Indeed, you have $(xy^{-1})^{n} = x^ny^{-n} = e$.

Answer 2

An equivalent formation of "$|x|$ divides $n$" is $x^n=e$ (why?).

Trivially, $e^n=1$, so $e \in H$.

Let $a,b \in H$, i.e. $a^n=b^n=e$. Then, $(ab^{-1})^n = a^n (b^{-1})^n = a^n (b^n)^{-1} = e e^{-1} = e$, so $ab^{-1} \in H$.

Answer 3

Notice that if $n=k|x|$ for some $k \in \mathbb{N}$, then $$x^n=x^{k|x|}=(x^{|x|})^k=e^k=e.$$ Conversely, if $x^n=e$, and we do division with remainder on $n$ by $|x|$, we get $n=k|x|+r$ for some positive $r<|x|$. So we see $$e=x^{k|x|+r}=x^{k|x|}x^r=ex^r=x^r,$$ so $r=0$, because $|x|$ is the smallest positive number $\alpha$ such that $x^\alpha=e$. So we see $$H=\{x \in G | x^n=e \}.$$ If we define $f:G \rightarrow G$ by $f(x)=x^n$, then since $G$ is abelian $(ab)^n=a^nb^n$ for all $a,b \in G$ so $f$ is a homomorphism, and $\ker f =H$, so $H$ is a subgroup.

Answer 4

Let $e$ be the identity of $G$.

Since $|e|=1|n$ , $e\in H$.

Hence $H$ is non-empty.

Let $a,b\in H $.

Since $b\in G$ and $G$ is a group, there exists $b^{-1}$.

Also $|b^{-1}|=|b|$.

Since $G$ is abelian ,$(ab^{-1})^n=a^nb^{-n}$ for all $n\in \mathbb{N}$.

Note that $a,b\in H$ $\Rightarrow$ $a^n=b^n=e$.

Hence $(ab^{-1})^n=a^n(b^n)^{-1}=e$ ,and $|ab^{-1}||n$.

Therefore $ab^{-1}\in H$ and $H$ is a subgroup of $G$.