# Prove a result on the size of the minimal set that generates a finite abelian group

I am asked to prove the following:

Let $G$ be a non-trivial, finite abelian group. Let $s$ be the smallest positive integer such that $G = \langle a_1,...,a_s\rangle$ for some $a_1,...,a_s \in G$. Show that $s$ is equal to the number $t$ in the relation: $$G \cong \mathbb{Z}_{m_1} \oplus\cdots\oplus \mathbb{Z}_{m_t}$$ where $m_i \mid m_{i+1}$ for $i=1,..,t-1$.

We know that such a list of integers ($m_1,...,m_t$) exists, and is unique by The Fundamental Theorem of finite abelian groups.

I tried approaching the problem as follows:

Let $\rho : \mathbb{Z}_{m_1} \oplus\cdots\oplus \mathbb{Z}_{m_t} \rightarrow G$ be an isomorphism, then clearly: $$G= \langle\ \rho(1,0,...0)\ ,\ \rho(0,1,...0)\ , ... ,\ \rho(0,0,...1)\ \rangle$$ , which implies $s\le t$. However I could not figure out how to show that $t \le s$.

• Very nicely written question AND informative title, and it's always good to see the effort you've put into the proof. (+1). – Namaste Jan 30 '14 at 14:16
• Consider the map $$\mathbb Z^s \to G \cong \mathbb{Z}_{m_1} \oplus\cdots\oplus \mathbb{Z}_{m_t}$$ which sends $(0,..,1,0,..,0)$ to $a_i$. The fact that this map is surjective should give you the other inequality. – zozoens Jan 30 '14 at 14:21
• @zozoens could you please elaborate? I can use the map you mentioned to show that $$\mathbb{Z}_{m_1} \oplus\cdots\oplus \mathbb{Z}_{m_t} \cong \mathbb{Z}_{k_1} \oplus\cdots\oplus \mathbb{Z}_{k_s}$$ where $k_i$ is the order of $a_i$, but I can't see how that helps. – SomeStrangeUser Jan 30 '14 at 14:50
• I believe you can always do invertible linear combinations on the $a_i$'s so that their orders are dividing one another. The unicity of the Fundamental Theorem then gives you the answer. – zozoens Jan 30 '14 at 15:24
• I admit I haven't seen this kind of things for a long time so I might be wrong. – zozoens Jan 30 '14 at 15:24

## 1 Answer

A clean way to do it consists perhaps in noting that if $p$ is a prime dividing $m_{1}$, then $G$ has a quotient group $Q$ isomorphic to $\mathbb{Z}_{p}^{t} = \mathbb{Z}_{p} \oplus \dots \oplus \mathbb{Z}_{p}$.

If in $$G \cong \mathbb{Z}_{m_1} \oplus\cdots\oplus \mathbb{Z}_{m_t}$$ a generator of the $\mathbb{Z}_{m_{i}}$ summand is $a_{i}$, then $$Q = G / \langle p a_{1}, \dots , p a_{k} \rangle.$$ In fact $$Q = \langle a_{1}, \dots , a_{k} \rangle / \langle p a_{1}, \dots , p a_{t} \rangle \cong \bigoplus_{i=1}^{t} \langle a_{i} \rangle / \langle p a_{i} \rangle \cong \mathbb{Z}_{p}^{t},$$ as the order of each $a_{i}$ is a multiple of $p$.

Even without appealing to the theory of vector spaces (see the comment by OP below), a subgroup of $Q$ generated by $k$ elements is easily seen to have at most $p^{k}$ elements, so $Q$ cannot be generated by less than $t$ elements.

Thus $G$, too, cannot be generated by less than $t$ elements, as any set of generators for $G$ will induce a set of generators for $Q$.

Now, as OP already noted, the isomorphism $$G \cong \mathbb{Z}_{m_1} \oplus\cdots\oplus \mathbb{Z}_{m_t}$$ shows that $G$ has indeed a set of generators of size $t$.

• I encountered this question in a textbook (Exercise 6.50, page 165), and it appears before the theories of rings or vector spaces are introduced, so I believe that the question should be answered without refering to those subjects. – SomeStrangeUser Jan 31 '14 at 22:20
• @SomeStrangeUser, in any case my approach reduces the question to groups of the form $Q = \mathbb{Z}_{p}^{t}$, which are easy to deal with: a subgroup of $Q$ generated by $k$ elements is easily seen to have at most $p^{k}$ elements, so... – Andreas Caranti Jan 31 '14 at 22:50
• Thanks you for your answer, but something still seems unclear to me. Could you perhaps clarify why $Q=G/\langle pa_1, ..., pa_k \rangle$ is isomorphic to $\mathbb{Z}_p^t$? – SomeStrangeUser Feb 1 '14 at 10:01
• @SomeStrangeUser, I have elaborated a little in my answer. – Andreas Caranti Feb 1 '14 at 11:30