Problem 5.2.11 (Minimum number of generators of a nontrivial finite abelian group of rank $t$)

Proof. Let $G$ be a nontrivial finite abelian group of rank $t$, so that

where $n_{i+1}\mid n_i$ for $1\le i<t$.

(a)

The section “Obtaining Invariant Factors from Elementary Divisors” shows that the rank $t$ of $G$ is the maximum of the rank of its Sylow subgroups.

(b)

Let $x_1,x_2,\dots ,x_t$ be generators of $Z_{n_1},Z_{n_2},\cdots ,Z_{n_t}$ respectively. If we identify $x_1\in Z_{n_1}$ with $(x_1,1,\dots ,1)\in Z_{n_1}\times Z_{n_2}\times \cdots \times Z_{n_t}$ (and so on), then $$G=⟨x_1,x_2,\dots ,x_t⟩.$$

Assume that $G=⟨y_1,y_2,\dots ,y_s⟩$. We must prove $t\le s$.

Let $\phi :G\to G$ defined by $\phi (x)=x^p$. By Exercises 7 and 8, $G∕\mathrm{im}(\phi )\simeq {(Z_p)}^t$, where $t$ is the rank of $G$ by part (a). In additive notations, there is an isomorphism

$$\psi :G∕\mathrm{im}(\phi )\to {𝔽}_p^t,$$

where $({𝔽}_p^t,+)$ is the additive group of the vector space $({𝔽}_p^t,+,\cdot )$.

Moreover, $G∕\mathrm{im}(\phi )=⟨\overline{x_1},\overline{x_2},\dots ,\overline{x_s}⟩$, where $\overline{x_i}=x_i\mathrm{im}(\phi )$. If $z_i=\psi (x_i)$, then ${𝔽}_p^t=⟨z_1,z_2,\dots ,z_t⟩$. The elements $(z_1,z_2,\dots ,z_s)$ generate the group $({𝔽}_p^t,+)$, therefore the vector space ${𝔽}_p^t$, of dimension $t$, is spanned by $(z_1,z_2,\dots ,z_s)$. This implies $t\le s$.

In conclusion, $G$ can be generated by $t$ elements but no subset with fewer than $t$ elements generates $G$.