# Powers of a group element only generate other group elements

I was reading some proof for the following theorem related to the order of an group element:

Theorem: Let $$x$$ be an element of a finite group $$(G, \circ)$$. Then $$x$$ has finite order.

Proof: Consider the list of consecutive powers of $$x$$:

$$..., x^{-1}, e, x^1, ...$$

The elements in this list cannot all be distinct, because they are all in $$G$$ [...]

I saw the bold part (all powers of $$x$$ are in $$G$$) claimed in various other proofs, but I never saw this derived.

I know it's true, but to me it's at least not an obvious fact.

I tried to come up with my own justification, although the proof turned out longer than I wanted:

Theorem: Assume a finite group $$(G, \circ)$$, then for some $$x \in G$$ of order $$n$$, the powers of $$x$$ are all elements of $$G$$.

Proof: We know $$x^2 = x \circ x = y$$ must be element of $$G$$, because the group $$G$$ is closed under $$\circ$$.

Also $$x^{-2} = z$$ must be also element of $$G$$, because it's the inverse of $$y$$.

We established $$y, z \in G$$. Composing each of them with $$x$$ creates a new power of $$x$$ and must also give a new element of $$G$$ by the closure property. This can be generalized in the following way:

If we pick any power $$x^k = x^{k-1} \circ x$$, then $$x^{k-1}$$ must be in $$G$$. Thus $$x^k$$ must be in $$G$$.

So the list of powers of $$x$$ can only contain elements of G.

In the proof I'm struggling to establish the general case from the $$x^2$$ case, so I'm unsure if this proof would be acceptable in its current form.

• It comes from the definition. The operator $\circ$ is a closed operation on the set. – Don Thousand Feb 22 at 23:31
• By mathematical induction. – Danny Pak-Keung Chan Feb 22 at 23:32
• Your key sentence is: 'because the group is closed under $\circ$': the operation on any two elements must yield a group element. – Berci Feb 22 at 23:33
• You can also get it with the pigeonhole principle – Michael Burr Feb 22 at 23:34

By definition, $${\circ}\colon G\times G\to G$$ is a map with codomain $$G$$, i.e., it maps everything to some element of $$G$$.
Since $$G$$ is finite, then $$S=\{x^n : n\in\mathbb Z\}$$ is also finite since it's a subset of $$G$$. Thus some $$x^n$$ are repeated; otherwise we could set up a bijection between $$\mathbb Z$$ and $$S$$ which would prove that $$G$$ is infinte.
Your proof is basically correct — you're showing that repeatedly applying $$\circ$$ to things in $$G$$ keeps you in $$G$$ — but this is quite obvious.