Cyclic finite groups and their direct product

Question

Let $C$ be a cyclic group with three elements. prove that $C \times C \times C$ can not be generated by two elements. I was thinking that showing that $C ^{3}$ is isomorphic to ${\mathbb{Z}_{3}} ^{3}$ will get me somewhere but I don't know how to proceed.

Answer 1

Suppose $a,b\in C\times C\times C$ are two nonidentity elements. Then the order of $a$ and $b$ is 3 because this is true of each nonidentity coordinate. The subgroup generated by $a$ and $b$ consists of all elements of the form $a^pb^q$ where $0\leq p,q\leq 2$. This subgroup has at most $9$ elements and hence cannot be the whole group, which has $27$ elements.

Answer 2

Uria, your original idea also works. $\mathbb Z_3$ is also a field, $(\mathbb Z_3)^3$ is a three-dimensional vector space over it, and it's subgroups are the same as its linear subspaces. Now you only have to refer to the fact that every vector space has a dimension (three in this case) and can not be generated by fewer elements.