1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It comes from the definition. The operator $\circ$ is a closed operation on the set. $\endgroup$ – Don Thousand Feb 22 at 23:31
  • $\begingroup$ By mathematical induction. $\endgroup$ – Danny Pak-Keung Chan Feb 22 at 23:32
  • $\begingroup$ Your key sentence is: 'because the group is closed under $\circ$': the operation on any two elements must yield a group element. $\endgroup$ – Berci Feb 22 at 23:33
  • 1
    $\begingroup$ You can also get it with the pigeonhole principle $\endgroup$ – Michael Burr Feb 22 at 23:34
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.