Short exact sequence of finite Abelian groups with the last group having odd order.

Question

We have a short exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$. And we assume that the order of $C$ is odd. Let $f$ be the map from $A$ to $B$.

One has to show that the map $\mathbb{Z}/2\mathbb{Z} \otimes_\mathbb{Z} A \rightarrow \mathbb{Z}/2\mathbb{Z} \otimes_\mathbb{Z} B$ induced by $f$ is an isomorphism.

My results so far:

I think that the map that is induced by $f$ is given by $\text{Id}\otimes f$. I have shown surjectivity of this map, where I used that an order of an element divides the order of the group ($C$ in this case).

Now I'm stuck at injectivity. I tried to prove that the kernel of this map is trivial. Where I wanted to use the property that a pure tensor $x\otimes y$ is zero if and only if $q(x,y) = 0$ for every $\mathbb{Z}$ bilinear map $q$.

Help is much appreciated.

Answer 1

For a $\mathbb{Z}$-module $M$ the map $M/2M\to\mathbb{Z}/2\mathbb{Z}\otimes_\mathbb{Z}M$, $m+2M\mapsto 1\otimes m$ is an isomorphism, which can be shown by using the universal property of the tensor product. So we can identify the above modules.

Use this for a diagram chase: Let $a+2A\in A/2A$ such that $f(a)+2B$ is trivial, i.e. there is an element $b\in B$ such that $2b=f(a)$. Now I don't want to spoil the fun and leave the rest to you.

Answer 2

Recall the well-known exercise: $$\mathbb{Z}/m\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/d\mathbb{Z}$$ where $d$ is the greatest common divisor of $m$ and $n$.

Suppose $$A = A_e\times \mathbb{Z}/x\mathbb{Z}$$ $$B = B_e\times\mathbb{Z}/y\mathbb{Z}$$ for odd integers $x,\,y$ (maybe 1) and $\left| A_e\right|=2^m$, $\left| B_e\right|=2^n$. By the hypothesis on the order of $C$, we immediately have $m=n$.

Note that $f$ restricts to $f\restriction : A_e\rightarrow B = B_e\times\mathbb{Z}/y\mathbb{Z}$ whose image lies in $B_e$ (compose $f\restriction$ with $\pi_2$), so that we obtain an injective map from $A_e$ to $B_e$. Since $\left| A_e\right|=\left| B_e\right|$, this is an isomorphism; in particular, $A_e$ and $B_e$ have the same number of elementary divisors.

Together with the mentioned exercise above, $\left|\mathbb{Z}/2\mathbb{Z} \otimes_{\mathbb{Z}} A\right|=\left|\mathbb{Z}/2\mathbb{Z} \otimes_{\mathbb{Z}} B\right|$. One more use of the pigeonhole principle on $1\otimes f$ ends the proof.