# Proving minimal number of generators for $\mathbb{Z}_n^k$ is $k$

It's very intuitive that the minimal size of a generating set for the group $\mathbb{Z}_n^k=\mathbb{Z}_n\times...\times\mathbb{Z}_n$ is exactly $k$. However, I was actually unable to find a simple proof of this trait (or even better - of a generalized theorem for a product of finite cyclic groups).

I know that for $n=2$ this can be easily proven since $\mathbb{Z}_2^k$ is a vector space. I also know one can use theorems related to free-abelian groups together with homomorphisms from such groups to finite abelian groups to prove the desired result, but it seemed to me like an overkill for a trait that appears to be very basic.

Are there such proofs that are more "elementary"?

• Your argument for $n=2$ works for any prime, and for any $n>1$ ${\mathbb Z}_n^k$ has a quotient ${\mathbb Z}_p^n$ for some prime $p$, – Derek Holt Sep 18 '18 at 8:00
• Can you please expand on the case of a general non-prime $n$? – Dean Gurvitz Sep 18 '18 at 8:12

Eventually I found a more elementary proof using combinatorics. The number of elements in $\mathbb{Z}_n^k$ is $n^k$, and the maximal order for each element is $n$. Because $\mathbb{Z}_n^k$ is abelian, the group generated by some elements $x_1,...x_m\in\mathbb{Z}_n^k$ can be written as the set: $$\{x_1^{a_1}*...*x_m^{a_m}|a_1,...,a_m\leq n\}$$ The size of this set is at most $n^m$, and therefore, for the set to be generating, it must contain at least $k$ elements. Since the standard Euclidean base is a generating set of size $k$, we can see that the minimal size of a generating set is indeed exactly $k$.

$\mathbb Z$ is not a vector space but rather a module over the commutative ring $\mathbb Z$. It is finitely generated as a product of finitely generated modules. A system of $k$ linear independent elements in $\mathbb Z$ then form a minimal generating set. Note that in general for modules you have to distinguish between a generating set and a basis. However, in your case it should be possible to repeat the proof that $k$ linear independent vectors form a basis for a vector space $V^k$ to prove that $\mathbb Z^k$ is generated by $k$ linearly independent vectors.

• Thank you, however, I am relatively unfamiliar with the language of modules and rings. I will further look into the linear algebra proof for a vector space to see if it helps. – Dean Gurvitz Sep 18 '18 at 8:07
• To define a module you just take the definition of a vector space (=Abelian over a field having scalar multiplication) and replace the field by a ring. Therefore, a module is an Abelian group over a ring having scalar multiplication. Easiest example: every module over a field is a vector space. Have a look in your favorite linear algebra textbook -- where did you really need that you are over a field? almost nowhere! usually, a ring suffices. This is why you can carry over a lot of linear algebra to modules, including things like generating sets. – James Sep 18 '18 at 8:27

Let $\{ C_1, \dots, C_k \}$ be a collection of cyclic groups s.t. $C_i = < a_i >$ and $|C_i| = n_i$, $\forall i = 1 \div k$; and define: $$C = C_1 \times \dots \times C_k$$ Then, it can be readily proven that: $$\tilde{a_i} = (e_1, \dots, a_i, \dots, e_k) \in C \> \land \> \tilde{C_i} \leq C, \forall i = 1 \div k$$ Therefore, due to the definition of Cartesian product, $C$ has $k$ generators: $\{ \tilde{a_1}, \dots, \tilde{a_k} \}$.

• But how would you prove that this is the minimal number of generators? – Dean Gurvitz Sep 18 '18 at 8:08
• Suppose you have an element $x = (x_1, \dots, x_k) \in C$, you can express each $x_i$ as a power of $a_i$; therefore, you can express $x$ as powers of $\tilde{a_1}, \dots, \tilde{a_k}$. Ergo, you don't need more generators. – AlbertCG93 Sep 18 '18 at 8:14
• Thank you, but unfortunately this still only proves an upper bound on the number of needed generators, and not a lower bound. – Dean Gurvitz Sep 18 '18 at 8:15
• I'm guessing you are considering the possibility of joining two groups together, let's say $\bar{C_1} \cong C_1 \times C_2$, and then having $C = \bar{C_1} \times C_3 \times \dots \times C_k$; is that it? – AlbertCG93 Sep 18 '18 at 8:21
• I know it isn't actually possible to do so, but I am looking for a proof for that. As I said, the minimality is a very intuitive trait to which I didn't find a simple proof. – Dean Gurvitz Sep 18 '18 at 8:25

I am not sure if you will like this argument, but it gives a more general result. Your question is a special case of the proposition below with $(R,I):=(\mathbb{Z},n\mathbb{Z})$. The proposition below, while not elementary, can be used to deal with the case where $(R/I)$ is not a finite ring, where your argument based on counting does not work. (This is not to say that your counting argument was not good. I like it very much.)

Proposition. Let $R$ be a commutative unital ring and $I\subsetneq R$ an ideal of $R$. Then, for a positive integer $k$, the $R$-module $(R/I)^k$ is finitely generated (as an $R$-module) and the minimum possible cardinality of a generating set of the $R$-module $(R/I)^k$ is $k$.

First, we know that $(R/I)^k$ is a finitely generated $R$-module,since the finite set $\left\{e_1,e_2,\ldots,e_k\right\}$ generates $(R/I)^k$, where $$e_j:=\left(0_{R/I},\ldots,0_{R/I},1_{R/I},0_{R/I},\ldots,0_{R/I}\right)\in (R/I)^k$$ in which $1_{R/I}$ appears at the $j$-th coordinate (and every other coordinate is $0_{R/I}$) for $j=1,2,\ldots,k$. Now, suppose that $m$ is the smallest positive integer such that $(R/I)^k$ is generated by some $$u_1,u_2,\ldots,u_m\in(R/I)^k$$ as an $R$-module. We then have $m\leq k$.

Let $J$ be a maximal ideal of $R$ containing $I$. Then, $J/I$ is a maximal ideal of $R/I$. Thus, $$F:=R/J\cong\frac{(R/I)}{(J/I)}$$ is a field. Let $\pi:(R/I)\to F$ be the canonical projection. Define the $R$-module homomorphism $\pi_k:(R/I)^k\to F^k$ to be the map sending $$\left(x_1,x_2,\ldots,x_k\right)\mapsto \big(\pi\left(x_1\right),\pi\left(x_2\right),\ldots,\pi\left(x_k\right)\big)$$ for all $x_1,x_2,\ldots,x_k\in R/I$. For $i=1,2,\ldots,m$, set $$v_i:=\pi_k\left(u_i\right)\in F\,.$$ We claim that $v_1,v_2,\ldots,v_m$ span the $F$-vector space $F^k$. Since $\dim_F\big(F^k\big)=k$, we conclude that $m\geq k$, and so $m=k$ must hold.

Write $\epsilon_j:=\pi_k\left(e_j\right)$ for $j=1,2,\ldots,k$. We only need to show that $$\epsilon_j\in\text{span}_F\left\{v_1,v_2,\ldots,v_m\right\}$$ for all $j=1,2,\ldots,k$. However, this is not difficult. Since the elements $u_1,u_2,\ldots,u_m$ generate the $R$-module $(R/I)^k$, there are coefficients $r_\nu^\mu\in R$, where $\mu\in\{1,2,\ldots,m\}$ and $\nu\in\{1,2,\ldots,k\}$, such that $$e_\nu=\sum_{\mu=1}^m\,r^\mu_\nu\,u_\mu\,.$$ That is, $$\epsilon_\nu=\pi_k\left(e_\nu\right)=\sum_{\mu=1}^m\,\pi\left(r^\mu_\nu\right)\,v_\mu$$ for $\nu=1,2,\ldots,k$. The claim is now proven.

P.S. You can modify my proof specifically for $R:=\mathbb{Z}$ and $I:=n\mathbb{Z}$, where $n>1$ is an integer as follows. Let $p$ be a prime natural number dividing $n$. Set $J:=p\mathbb{Z}$ (so $F=\mathbb{F}_p$ is the finite field of order $p$). Everything follows in the same manner.

• As I replied to James, I am unfortunately not fluent enough in the language of rings and modules to properly appreciate this answer. Is this basically a conversion of the proof of basis size for vector spaces to modules over rings? – Dean Gurvitz Sep 18 '18 at 10:21
• See the post scriptum (P.S.) if you don't want to deal with rings and modules. (The term "(unitary) $\mathbb{Z}$-modules" is just an alias of "abelian groups.") – Batominovski Sep 18 '18 at 10:23