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Explore the Back-Story**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}).

**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}).

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