Prove $H=\lbrace x \in G: x^n =e \rbrace$ is a subgroup of group G.

Question

Prove $H=\lbrace x \in G: x^n =e \rbrace$ is a subgroup of group G if n is a fixed integer. (G is abelian)

I understand the basis of this proof however my professor has asked that I make a small revision to my proof. I proved that H is closed, nonempty and contains the identity however I am having trouble proving the existence of inverses. My for inverses argument is as follows...

Let $x \in H$, then $x^n=e$ and $$((x^{-1})^n)=((x^n)^{-1})=e^{-1}=e$$

I have been asked to prove why $((x^{-1})^n)=((x^n)^{-1})$. Any suggestions?

Answer 1

Shoe-sock theorem comes to mind.

That is

$$(xy)^{-1}=y^{-1}x^{-1}$$

Which for abelian groups is even nicer to us.

Answer 2

Notice that $(x^{-1})^n$ is the $n$th power of $x^{-1}$, while $(x^n)^{-1}$ is the inverse of $x^n$. Those aren't a priori identical objects. To show that $(x^{-1})^n$ is the inverse of $x^n$, show that it satisfies the characteristic property of $x^n$: after multiplying by $x^n$, you get $e$.

Now $$ (x^{-1})^nx^n = \underbrace{x^{-1} x^{-1} \cdots x^{-1}}_{\text{$n$ times}} \cdot \underbrace{x \cdot x \cdots x}_{\text{$n$ times}} $$ You can see that the $x^{-1} x$ in the middle is $e$, so it drops out. But then the same thing happens with the other $n-1$ copies of $x^{-1}$ and $x$, in pairs. So in the end you are left with $e^n = e$. Thus $(x^{-1})^n = (x^n)^{-1}$.

Answer 3

Let $x \in H$. Then $x^n = e$ for some $n$.

Now since $x^n=e$ for some $n$ this implies that $x^{-1}x=e$ by group properties.

I claim that $x^{-1} \in H$

Proof:

$x^{-1}x=e \to (x^{-1}x)^n=e^n=e \to (x^{-1})^nx^n=e$ since $G$ is abelian

But since $x^n=e$, then $(x^{-1})^n=e$

So $x^{-1} \in H$

Answer 4

I don't know if you have encountered this yet, but for any $x\in G$, we have that $|x| = |x^{-1}|$. This implies that $(x^{-1})^n=e=(e)^{-1}=(x^n)^{-1}$, which is to say tha $(x^{-1})^n=(x^n)^{-1}$.

I hope this helps.

Edit:

I'm assuming that in this context $n$ is non-negative. However, the proof for $n< 0$ is similar, so WLOG assume $n\geq 0$.

Note $(x^n)^{-1}=e^{-1}=e$, so the goal is to show $(x^{-1})^n=e$: Since only $|e|=1$, we will assume that $n>1$. Since $x\cdot x^{n-1}=e$, $x=x^{-(n-1)}=(x^{-1})^{n-1}$ (in general, whenever $k \geq 0$, we can say that $g^{-k}=(g^{-1})^k$ for all $g\in G$).

Next, note that $x\cdot x^{-1} = (x^{-1})^{n-1}\cdot x^{-1}=(x^{-1})^{n}$. Since $x\cdot x^{-1} = e$, it follows that $(x^{-1})^{n}=e$. Therefore, $x^{-1} \in H$.

Answer 5

The set in question is $$H = \{ x : x^n = e \}.$$

Certainly this set contains the identity $e$.

Is it closed under the group operation? Let $x,y \in H$, then $(xy)^n = x^n y^n = e^2 = e$.

Does it contain the inverse? If $x^{-1}$ is the inverse, then $(x^{-1})^n = (x^n)^{-1}= e^{-1} = e.$

Why is $(x^{-1})^n = (x^n)^{-1}$? This can be seen via induction. The base case is clear, now assume the case is true for $n$. Then consider the $n+1$ fold product

\begin{align} x^{-1} \underbrace{x^{-1} \dots x^{-1}}_{n} &= x^{-1}(x^n)^{-1} \quad (IH)\\ &=(x^1)^{-1}(x^n)^{-1} \\ &=(x^nx^1)^{-1}\\ &=(x^{n+1})^{-1}. \end{align}

Addendum: If one regards $\phi(x) = x^n$, then set in question is the kernel of $\phi$, which of course is a subgroup of $G$.