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"?
Best Answer
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 $k$.