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.
Best Answer
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.