Fix a natural number $n$. Let $A$ be a 2-torsion finite abelian group . Let $f: A\to (\Bbb{Q}/\Bbb{Z})^n$ be an arbitrary group homomorphism. How can I prove $image(f)\le 2^n$?
My try: From structure theorem of finite abelian group , $A\cong (\Bbb{Z}/2\Bbb{Z})^r$ for some integer $r$.Take Pontryagin dual of $f^*$. Then, $f^*: (\hat{\Bbb{Z}})^n=(\prod_p \Bbb{Z}_p)^n\to A^*\cong (\Bbb{Z}/2\Bbb{Z})^r$. By the property of Pontryagin dual, $\#image(f)=\#image(f^*)$.$image(f^*)\cong (\prod_p \Bbb{Z}_p)^n/(\prod_p p^m\Bbb{Z}_p)^n\cong (\Bbb{Z}/p^m\Bbb{Z})^n$.Bút thí group should be 2-torsion, thus $p^m\le 2$. So the result follows.
But my try should be overtaking, can we prove this without using Pontryagin dual or is it standard way to use Pontryagin dual in such a situation ?
The $2$-torsion part of $\mathbb{Q}/\mathbb{Z}$ is $\left\langle \frac{1}{2}+\mathbb{Z}\right\rangle\cong\mathbb{Z}/2\mathbb{Z}$.
Indeed, if $\frac{a}{b}+\mathbb{Z}$ has order dividing $2$, with $\gcd(a,b)=1$, then $\frac{2a}{b}\in\mathbb{Z}$, so $b\mid 2a$; hence $b\mid 2$, so either $b=1$ and $\frac{a}{b}+\mathbb{Z}=0+\mathbb{Z}$, or $b=2$ and $\frac{a}{b}+\mathbb{Z}=a\left(\frac{1}{2}+\mathbb{Z}\right)$.
An element $(a_1,\ldots,a_n)$ of $A_1\times\cdots\times A_n$ has order dividing $2$ if and only if $a_i$ has order dividing $2$ in $A_i$.
So the image of $f$ lies in $(\frac{1}{2}+\mathbb{Z})^n\cong (\mathbb{Z}/2\mathbb{Z})^n$. Therefore, the image has order dividing $2^n$. This does not require $A$ to be finite (though any $2$-torsion abelian group is a vector space over $\mathbb{F}_2$, so it is isomorphic to a direct sum of copies of $\mathbb{Z}/2\mathbb{Z}$).