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Let $A$ be a finite subset of an abelian group $G$. For an integer $n$, define the dilate $n \cdot A = \{na \mid a \in A\}$ and, given another $B \subseteq G$, the sumset $A + B = \{a + b \mid a \in A, b \in B\}$.

I would like to show that $$ \frac{|n \cdot A + (nm) \cdot A|}{|n \cdot A|} = \frac{|A + m \cdot A|}{|A|} $$ for any $n,m \in \mathbb{Z}$.

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up vote 3 down vote accepted

Note that $nA+(nm)A = n(A+mA)$. To verify this, if $x\in nA+(nm)A$, then there exists $a,a'\in A$ such that $x = na + nma' = n(a+ma')$; and $a+ma'\in A+mA$. Conversely, $n(a+ma') = na+nma'\in nA+(nm)A$, if $a,a'\in A$.

So we are asking whether $$\frac{|n(A+mA)|}{|nA|} = \frac{|A+mA|}{|A|}.$$

If $\langle A\rangle$ has no $n$-torsion, then the result follows since multiplication by $n$ is one-to-one (if $na = na'$ then $n(a-a')=0$, so $a-a'\in\langle A\rangle$ is $n$-torsion, so $a=a'$; the same holds for $A+mA\subseteq \langle A\rangle$).

However, if $A$ has $n$-torsion, then you are in trouble. Consider for example $n=5$, $m=3$, $G=\mathbb{Z}/10\mathbb{Z}$, $A=\{1,2\}$. Then $A+3A = \{1,2\}+\{3,6\} = \{4,5,7,8\}$, so the right hand side is $2$. On the other hand, $5(A+3A) = 5\{4,5,7,8\} = \{0,5\}$ and $5A = \{0,5\}$, so the left hand side is $1$.

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I knew I was home free if there was no $n$-torsion, but I hadn't managed a counter-example. Thanks for the sanity check. –  Hans Parshall Jul 25 '12 at 19:58
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I'll assume $n$ is a non-zero integer and $G=\mathbb{Z}$; the result is not true for general $n,A$.

Multiplication by $n$ is an injection $\mathbb{Z}\to\mathbb{Z}$, so $|n\cdot X|=|X|$ for any set $X\subset\mathbb{Z}$. The equality follows by applying this to $X=A$ and $X=A+m\cdot A$.

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