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Is it true if $A$ is a finite, abelian group and $m$ is some integer relatively prime to the order of $A$, then the map $a\mapsto ma$ is an automorphism?

It's left as an exercise in some course notes, but I cannot verify it.

In particular it is used to prove that Hall complements exist for abelian Hall subgroups of finite groups.

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    If you're talking about groups, then notice that the map $a \mapsto ma$ for some integer $m$ is in general only a homomorphism if $A$ is abelian.2012-10-19

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Hint: since $\gcd(m,\#A)=1$, you can find an integer $d$ such that $dm\equiv1\pmod{\#A}$. Show that the map $a \mapsto da$ is the inverse of $a \mapsto ma$.

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Hint: To prove injectivity (and therefore surjectivity), suppose that $ma=mb$. Then $m(a-b)=0$. So $a-b$ has order a factor of $|m|$. But the order of $a-b$ must divide the cardinality $\# A$ of $A$. Since $m$ and $\#A$ are relatively prime, it follows that $a-b$ has order $1$, that is, $a-b=0$.

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    You may assume $b=0$ and therefore simplify the proof (injective <-> kernel zero).2012-10-20
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Hint $\ $ Let $\rm\:n = |A|.\:$ By Lagrange and Bezout, $\rm\: na, m\,a= 0\:\Rightarrow\: (n,m)a = 0.\:$ Here $\rm\:(n,m) = 1.$