Torsion in abelian groups

Question

If $A$, $B$ and $C$ are finite abelian groups that obey the following exact sequence $$A\rightarrow B\rightarrow C\rightarrow1$$ and $$A[m]:=\{a\in A:a^m=1\}$$ is the following inequality true or false, why? $$|B[m]|\leq |A[m]||C[m]|$$

I know that it is false when I take away the finite condition. If it is false, is there some other nice upper bound on $|B[m]|$?

One way to start is by restricting attention to $p$ primary invariant, but I am not sure how to proceed. Maybe taking the quotient of $A[p]$ and $f(A[p])$ will help where $f:A\rightarrow B$ is the homomorphism in the exact sequence, and then induction, but I am not sure if the details follow.

Answer 1

The functor $F(M) = M\otimes \mathbb{Z}_m = M/mM$ is right-exact. Furthermore, if $M$ is finite, then $\#mM = \#(M/M[m])$ and $\#M[m] = \#F(M)$. The exact sequence \begin{align*} F(A) \to F(B) \to F(C) \to 1 \end{align*} then implies \begin{align*} \#F(C) = \frac{\# F(B)}{\# \ker F(B) \to F(C)} = \frac{\#F(B)}{\# \operatorname{im} F(A) \to \#F(B)} \geq \frac{\#F(B)}{\#F(A)}; \end{align*} that is, $\#F(B) \leq \#F(A) \#F(C)$.

Answer 2

Applying the exact contravariant functor $\operatorname{Hom}(-,\mathbb{Q}/\mathbb{Z})$ to your exact sequence yields an exact sequence $$1\to C\to B\to A$$ since $\operatorname{Hom}(X,\mathbb{Q}/\mathbb{Z})\cong X$ for any finite abelian group $X$. But now the conclusion is trivial, since the maps in this exact sequence restrict to a short exact sequence $$1\to C[m]\to B[m]\to D\to 1$$ where $D$ is the image of $B[m]$ under the map $B\to A$ and in particular is a subgroup of $A[m]$, so $|B[m]|=|C[m]||D|\leq|C[m]||A[m]|$.