Claim: Let $M$ be a $R$-module ($R$ is an integral domain) and $p \in R$ be a prime. Suppose there exists non-empty finite subsets $B$ and $C$ of $M \backslash\{0\}$ such that $M= \bigoplus_{m \in B} Rm=\bigoplus_{n \in C}Rn$ and for each $x\in B \cup C, \text{Ann}(x)=\{p^{k_x}\},$ for some $k_x\in \mathbb{Z}^{+}.$ Then there is a bijection $\sigma:B \to C$ such that $\text{Ann}(m)=\text{Ann}(\sigma(m)), $ for all $m \in B.$
Proof: Let $B^{\prime}=\{m \in B\ | \ \text{Ann}(m)=(p)\}.$ Then, $pM=\bigoplus_{m \in B \backslash B^{\ \prime}} R(pm)$ and thus $M/pM \cong \bigoplus_{m^{ \ \prime} \in B^{\ \prime}}Rm^{ \ \prime} \oplus \ \bigoplus _{m \in B \backslash B^{\prime}}Rm/R(pm)$ as $R/(p)$ modules and the latter has basis $\{m^{\prime} \in B^{\ \prime}\} \cup \{m+R(pm) \ | \ m\in B\backslash B^{\ \prime}\},$ so that $M/pM$ is free as $R/(p)$ module of rank $|B|.$ Prove by induction on $N=\text{min}\{k \in \mathbb{Z}^{+} \ | \ p^km=0, \forall m \in M\}.$ If $N=1,$ then $B=B^{\ \prime}$ and $C= C^{\ \prime},$ any bijection $\sigma:B \to C$ will do. For $N>1,$ since $\text{Ann}(pm)=(p^{k-1})$ if $\text{Ann}(m)=(p^k),$ with $k \in \mathbb{Z}^{+},$ we see by induction(applied to $pM$) that there is a bijection $\sigma^{\ \prime}:B \backslash B^{\ \prime} \to C \backslash C^{\ \prime}$ such that $\text{Ann}(pm)=\text{Ann}(p\sigma^{\prime}(m)),$ for all $m \in B \backslash B^{\ \prime}.$ So $\text{Ann}(m)=\text{Ann}(\sigma^{\prime}(m)),$ for all $m \in B \backslash B^{\ \prime}.$ Let $\sigma:B \to C$ be any bijection extending $\sigma^{\prime}...$
This proof is quoted from my lecture notes. I don't understand how is Induction being used here. What is the Induction hypothesis for example? Could someone advise me please?
Thank you.