Let $G$ be a finite abelian group of rank $t$.
(a) Prove that the rank of $G$ is the maximum of the ranks of $G$'s Sylow subgroups.
(b) Prove that the smallest subset of $G$ generating $G$ is of order $t$. [One way of doing this is by combining part (a) with exercise 7.]
Proof: (a) Obtain $G$'s invariant factor decomposition, and let $p$ be a prime dividing $n_s$, the order of the smallest cyclic factor. By 5.1.4, we have the invariant factor decomposition of $P∈\text{Syl}_p(G)$ is of rank $t$.
(b) We know $G$ is generated by the $t$ generators of each of the cyclic factors of $G$, so existence is not an issue. Assume that $G = < S >$, where $| S | = s < t$.
Obtain $P$ of rank $t$ by part (a), and let $φ$ be the $p$-power map. By 5.2.7(a), we have $G/\text{im}~φ ≅ \text{ker}~φ$. Prove that $\text{ker}~φ = \text{ker}~φ|_P$. ($\supseteq$) If $x∈\text{ker}~φ|_P$, then definitionally $x^p=1$ and $x∈ P ≤ G$. ($\subseteq$) Assume $x^p=1$ but $x \not ∈ P$. Since every $p$-subgroup is contained within a Sylow $p$-subgroup, and $P$ is the unique Sylow $p$-subgroup of $G$ by the abelian quality of $G$, we have $< x > ≤ P$, and thus $x∈P$. Therefore $\text{ker}~φ = \text{ker}~φ|_P$. By part (a), we have that $\text{ker}~φ|_P ≅ E_{p^t}$, so that $G/\text{im}~φ ≅ E_{p^t}$, and in particular - as a consequence of the former being generated by the natural projection image of $S$ - the latter is generatable by a subset of fewer than $t$ elements.
For a generating subset $K \subseteq E_{p^t}$ and $| K | = n < t$, we have every element of $E_{p^t}$ is rewritable as $k_1^{\alpha_1}k_2^{\alpha_2}...k_n^{\alpha_n}$. Since every nontrivial element of the elementary abelian group is of order $p$, we have an upper bound of $p^n$ distinct elements writable in this fashion, a contradiction since $p^n < p^t = |E_{p^t}|$. $\square$
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