As a ZZ-module, the p-primary components of a finite abelian group are its Sylow subgroups

Let $M$ be a finite abelian group of order $a = \prod p_i^{k_i}$ and consider $M$ as a $\mathbb{Z}$-module in the natural way. Prove that $M$ is annihilated by $(a)$, that the $p_i$-primary component of $M$ is the (unique) Sylow $p_i$-subgroup of $M$, and that $M$ is isomorphic to the direct product of its Sylow subgroups.

Certainly for all $m \in M$, $a \cdot m = 0$. Thus $(a) \subseteq \mathsf{Ann}_\mathbb{Z}(M)$. Since $\mathbb{Z}$ is a principal ideal domain, by this previous exercise, $M = \bigoplus (q_t)M$, where $q_t = \prod_{i \neq t} p_i^{k_i}$.

We claim that each $(q_t)M$ is a Sylow subgroup of $M$. To that end, note that $p_t^{k_t} \cdot (q_t)M = 0$; in particular, every element of $(q_t)M$ has $p_t$-power order. Conversely, if $m \in M$ has $p_t$-power order, then $m \in \mathsf{Ann}_M(p_t^{k_t}) = (q_t)M$. Hence the $(q_t)M$, that is, the $p_t$-primary components, are Sylow subgroups of $M$. Since $M$ is abelian, there is a unique Sylow $p_t$-subgroup for each $p_t$; thus $M = \bigoplus (q_t)M$ is the internal direct sum of its Sylow subgroups. Since the direct sum is finite, $M$ is in fact the internal direct product of its Sylow subgroups.

Post a comment or leave a trackback: Trackback URL.