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If $(G,\ast)$ is a group, $x,y \in G$, $x\ast y = y\ast x$, and $\gcd(o(x),o(y)) = 1$, then $o(xy) = o(x)o(y)$.

$o$ denotes the order.

I know how to begin getting details from what is given, $o(x) = x^m = e$, when $m$ is minimal.

$o(y) = y^k = e$, when $k$ is minimal.

In other words, $o(x) = m, o(y) = k, o(xy) = p$ I need to show that $p=mk$

I need to prove that $(xy)^p = x^m \cdot y^k, $when $p$ is minimal.

How to continue from here, no ideas.

Ilan Aizelman WS
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    So you want to prove that $p=mk$. Maybe try to prove that $p|mk$ and $mk|p$. You can prove that $p|mk$ by proving that $(xy)^{mk}=e$ and by the minimality of $p$, $p|mk$ follows. Now is it the case that $(xy)^{mk}=e$? Check it yourself. I didn't think about $mk|p$, but it should work around the same idea. – Git Gud Jan 08 '17 at 11:24

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You haven't used the fact that $x$ and $y$ commute yet!

Let $p = o(xy)$. Then

$$e = (xy)^p = \underbrace{(xy)(xy)\dots (xy)}_{p \text{ times}} = x^py^p.$$ So you also get that $$e = (xy)^{pm} = \underbrace{(xy)(xy)\dots (xy)}_{pm \text{ times}} = x^{pm}y^{pm} = (x^m)^py^{pm} = e^py^{pm} = y^{pm}.$$ Since $k$ is the order of $y$, we then have that $pm$ is a multiple of $k$, or thus $k|pm$. But $\gcd(m,k) = 1$ so $k|p$. Similarly, taking the power $pk$ gives $m|p$, and since $\gcd(m,k) = 1$, you know that $mk | p$.

On the other hand, you have that $$(xy)^{mk} = \underbrace{(xy)(xy)\dots (xy)}_{mk \text{ times}} = x^{mk}y^{mk} = (x^m)^k(y^k)^m = e^ke^m = e.$$ Hence $p | mk$.

But now we have that $mk|p$ and $p|mk$, so $mk = p$.

sTertooy
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