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Take an Abelian group G, and let $g,h \in G$ with $o(g) = x$ and $o(h) = y$, prove that if $x,y$ are coprime, then $o(gh) = xy$

(o(g) denotes the smallest integer x s.t. $g^x = e$)

My (bad) attempt:

since $o(g) = x$ then $o(g) | x$, similar $o(h) | y$

Now let $o(gh) = k$ for some integer k,

then since $x,y$ are coprime $o(g) | o(h)o(gh)$ implies $o(g) | o(gh)$, similarly $o(h) | o(gh)$ i.e. $k = z_1 x$ and $k = z_2 y$, so the only way for this to be possible is if $z_1 = y$ and $z_2 = x$ and we're done

is this the right approach? Could someone correct me somewhere? Also, why does G have to be Abelian?

Warz
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  • I think you mean that $o(g)$ denotes the smallest (not highest) positive integer $x$ s.t. $g^x=e$. – JiK Feb 12 '14 at 21:35
  • To me it's unclear how you get $o(g)|o(h)o(gh)$ in the first place. – JiK Feb 12 '14 at 21:37
  • @JiK The idea was that if you consider the product $o(h)o(gh)$ then you know that since $o(g)$ and $o(h)$ are coprime then $o(g)|o(gh)$ , we proved this in the previous part of the question so I figured I would have to use it somewhere. – Warz Feb 12 '14 at 22:37

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By abelianity: $$(gh)^{xy}=(g^x)^y(h^y)^x=e$$ thus $o(gh)|(xy)$. Now let $(gh)^n=e$, then $g^n=h^{-n}$, thus $h^{-nx}=g^{nx}=e$ hence $y\mid nx$ thus $y\mid n$ because $y$ is coprime with $x$. Similarly, $g^{ny}=h^{-ny}=e$, hence $x\mid ny$ that's $x\mid n$. Consequently, $xy\mid n$.

Fabio Lucchini
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