**Question:**

How many subgroups does the group \(\mathbb{Z\times Z}\) have?

A. 1

B. 2

C. 3

D. infinitely many.

**Discussion:**

How many subgroups does \(\mathbb{Z}\) have? Well, for every \(m\ge 0\) we have a subgroup \(m\mathbb{Z}\).

And these are all distinct. So there are infinitely many subgroups of \(\mathbb{Z}\).

So how many subgroups does \(\mathbb{Z}\times \{\bar{0}\}\) have? Again, for every \(m\ge 0\) we have a subgroup \(m\mathbb{Z}\times \{\bar{0}\}\). That is again infinitely many.

Since \(\mathbb{Z}\times \{\bar{0}\}\) is a subgroup of \(\mathbb{Z\times Z}\) we have **infinitely many** subgroups of \(\mathbb{Z\times Z}\).