# Thread: Meaning of a theorem

1. ## Meaning of a theorem

This is a simple theorem I have read online:

Let G be a group and let $\displaystyle g, h \in G$ be commuting elements of finite orders m, n, respectively,
where $\displaystyle gcd(m,n)=1$. Then $\displaystyle (g) \cap (h)$ is the trivial subgroup.
With the following proof:
Let $\displaystyle x \in (g) \cap (h)$. Then |x| is a divisor of |g| and of |h|, and so |x| is a divisor of
gcd(|g|,|h|)=1. Thus x is the identity of G.
The question is: What does "commuting element" mean? If it's the ordinary meaning (Like in ab=ba).
What is the role in the proof and why cannot the above theorem be proven for non-commuting elements?

2. ## Re: Meaning of a theorem

Your definition (ab=ba) of a commuting element does hold here.

It is important as elements don't have to commute inside a group. This proof only holds when elements do commute.

3. ## Re: Meaning of a theorem

the hypothesis that g and h commute is not needed, here. <g>∩<h> is a subgroup of <g> and <h>. by lagrange, its order is a common divisor of <g> and <h>, and thus of |g| and |h|.

but this means that |<g>∩<h>| divides gcd(|g|,|h|) = gcd(m,n) = 1, so |<g>∩<h>| = 1, that is: <g>∩<h> is trivial.

the condition that gh = hg may be important, however, in something else that the uses the proof, or that the proof builds upon. without knowing the context, i cannot say for sure.