$\#imagef=\#imagef^*$, order of image of dual and image of original homomorphism is the same

Question

Let $M$ be an abelian group Let $M^*=Hom_{\Bbb{Z}}(M,\Bbb{Q}/\Bbb{Z})$ be pontryagin dual of $M$.

Let $M$ be a finite group. Let $M'$ be an abelian group. Let $f:M\to M'$ be a homomorphism of abelian group. Let $f^*: M'^*\to M^*$ be a map defined by $g\to f・g$. Then, I want to prove $\#imagef=\#imagef^*$.

I know $\#M=\#M^*$. I'm having difficulty solving this problem because I have no idea of how to relate $imagef$ and $Imagef^*$. What is a tactics to tackle this kind of problem?

I want to solve this problem on my own, but I can't take the first step. I would be very happy if you could give me a hint, guidance, or advice.

Answer 1

Hint: Let $K=\ker f$ and consider the Pontryagin dual of the exact sequence $0\to K\to M\to M'$.

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Since Pontryagin duality is an exact functor, the dual sequence $$(M')^*\to M^*\to K^*\to 0$$ is also exact. Now just observe that these exact sequences give $|\operatorname{im} f|=|M|/|K|$ and $|\operatorname{im} f^*|=|M^*|/|K^*|$. In fact, more strongly, if $I=\operatorname{im} f$ then there is a short exact sequence $0\to K\to M\to I\to 0$ and dualizing this shows that $I^*$ is the kernel of $M^*\to K^*$, i.e. the image of $f^*$, so there is actually a canonical isomorphism $(\operatorname{im} f)^*\cong \operatorname{im} f^*$.