I have $G$ a finite abelian group of order $n$, an application $[m]: G \rightarrow G$ with $[m](g) = mg$ and $gcd(n,m)=1$. I want to prove that $[m]$ is surjective. I know there exist $s, t \in \mathbb{Z}$ such that $ns + mt = 1$, but I am not sure how to find the right $g \in G$ such that $[m](g) = mg = y$. So far I found that $y = 1y = (ns + mt)y = (ns + m(t-1) + m)y$, but it doesn't seem to help me that much. Can you help?
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What do you know about multiplication by $n$ in an abelian group of order $n$? – Thorgott Nov 09 '21 at 01:16
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@Thorgott I think $G$ is cyclic because $(n,m) = 1$. Is this what you meant? This is where I am confused – Patrick Nov 09 '21 at 01:18
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Same as in the dupe (except you're using additive notation, so $n$'th powers translate to multiples of $n)\ \ $ – Bill Dubuque Nov 09 '21 at 02:08
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You are actually very close to the solution.
Note that $ng = 0$ for any $g \in G$, by Lagrange's theorem. Together with Bezout's identity $ns + mt = 1$, we get $$[m](tg) = mtg = mtg + nsg = g$$ which shows that $g$ is in the image of $[m]$. This being true for any $g \in G$, the map $[m]$ is surjective.
Notice that we didn't use the assumption that $G$ is abelian. The result is true for general finite groups.
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Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Nov 09 '21 at 01:53