Show that the group Z/mZ × Z/mZ × Z/mZ, m > 1, can be generated by three elements and not two?
Showing it can be generated by three isn't bad at all. But how about showing that it can't be generated by two? I'm thinking about the old contradiction route of assume the opposite and find a contradiction, but this proof if trickier than its analogue in linear algebra, because with Z/mZ we have zero divisors...
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This is easily accomplished with a counting argument.
Suppose that {(a,b,c), (d,e,f)} is a Z-linearly independent subset of Z/mZ × Z/mZ × Z/mZ.
Observe that the set of all Z-linear combinations of these two elements
{r(a,b,c) + s(d,e,f) : r, s in Z} has at most m^2 distinct elements.
(This is where I'm using the finiteness of Z/mZ to our advantage.)
On the other hand, Z/mZ × Z/mZ × Z/mZ has m^3 elements. Needless to say,
{r(a,b,c) + s(d,e,f) : r, s in Z} does not generate all of Z/mZ × Z/mZ × Z/mZ.
I hope this helps!
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