## The rank of a finite abelian group is the maximum rank of its Sylow subgroups

Let $G$ be a nontrivial finite abelian group of rank $t$.

1. Prove that the rank of $G$ is the maximum of the ranks of its Sylow subgroups.
2. Prove that $G$ can be generated by $t$ elements, but that no subset of fewer than $t$ elements generates $G$.

1. Suppose (via FTFGAG) $G$ has type $(n_1,n_2,\ldots,n_t)$, where $n_i|n_j$ for $i \geq j$. Then without loss of generality, $G = Z_{n_1} \times \cdots \times Z_{n_t}$. By a previous exercise, every Sylow $p$-subgroup of $G$ is of the form $P = P_1 \times \cdots \times P_t$, where $P_i \leq Z_{n_i}$ is a Sylow $p$-subgroup. Again by FTFGAG, the type of $P$ is $(|P_1|, \ldots, |P_s|)$, where $s$ is maximal such that $P_s \neq 1$. In particular, the rank of $P$ is at most $t$. Moreover, if $p$ divides $n_t$, then $s = t$ and the rank of $P$ is $t$. Thus the rank of $G$ is the maximum of the ranks of is Sylow subgroups.
2. We begin with a lemma.

Lemma: The elementary abelian group $E$ of order $p^t$ is not generated by any subset of fewer than $t$ elements. Proof: Let $S \subseteq E$ be a subset consisting of $k < t$ elements. Now because $E$ is commutative and every element has order $p$, every element of $E$ can be written in the form $s_1^{a_1} s_2^{a_2} \cdots s_k^{a_k}$, where $0 \leq a_i < p$. Thus there are at most $p^k < p^t$ elements in $E$, a contradiction. $\square$

Again without loss of generality, $G = Z_{n_1} \times \cdots \times Z_{n_t}$, where $Z_{n_i} = \langle x_i \rangle$. Denote by $e_i \in G$ the element whose $i$th component is $x_i$ and whose other components are 1. Clearly $G = \langle e_1, e_2, \ldots, e_t \rangle$. Thus $G$ is generated by $t$ elements.

Suppose now that $G$ is generated by a subset $S$ containing $k < t$ elements. By part (1), there is a prime $p$ such that the (necessarily unique) Sylow $p$-subgroup $P$ of $G$ has rank $t$. Let $\varphi$ denote the $p$-power map on $G$; by a previous exercise, $G/\mathsf{im}\ \varphi \cong Z_p^t$. But then $Z_p^t$ is generated by $k < t$ elements, a contradiction of the lemma.

• will  On November 3, 2011 at 6:59 pm

Hi,

Thanks for the incredibly useful site!

A question about part 1:
Why must there necessarily be a Sylow p-subgroup such that p divides n_t? That is, why can it not be the case that the rank of the group is greater than that of any of its Sylow p-subroups?

• will  On November 3, 2011 at 7:11 pm

Just answered my own question—

It follows because there is at least one Sylow p-subgroups for every prime p, and the last invariant factor of G, n_t, is either a prime or a product of primes.