# The $p^k$-rank of a subgroup is no greater than the $p^k$-rank of the group.

Recently I was given a handout containing (roughly) the following text:

Let $A$ be a finite abelian group, and $p^k$ a prime power. The $p^k$-rank of $A$ is defined to be $$\operatorname{rank}_{p^k}(A):=\operatorname{dim}_{\mathbb{F}_p}(p^{k-1}A/p^kA).$$ It is immediate from the definition that for any subgroup $B\subseteq A$ and any prime power $p^k$ $$\operatorname{rank}_{p^k}(B)\leq\operatorname{rank}_{p^k}(A).$$

However, I do not see how this follows from the definition. Any suggestions?

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For the sake of simplicity we consider only the case $k=1$. We have to show $\dim_{\mathbb F_p}(B/pB) \leq \dim_{\mathbb F_p}(A/pA)$. Note that this is equivalent to $|B/pB| \leq |A/pA|$. Consider the exact sequence $$0 \to {}_pA \to A \to pA \to 0$$ where $A \to pA$ is multiplication by $p$ and ${}_p A = \{ a \in A : pa = 0 \}$ is its kernel. The exactness implies $|A| = |{}_pA| |pA|$ and similarly $|B| = |{}_pB| |pB|$. Since ${}_p B \subseteq {}_p A$, we get $$|B/pB| = |B|/|pB| = |{}_pB| \leq |{}_p A| = |A|/|pA| = |A/pA|.$$