## Mr.Math 4 years ago A problem in group theory. Prove that a group has exactly $$3$$ subgroups if and only if it's a finite cyclic group of order $$p^2$$.

1. anonymous

Well, we know that the order of a subgroup must divide the order of the group. So, if we have a finite cyclic group of order $$p^2$$, its subgroups will necessarily have order $$1, p, \text{or } p^2$$. We can prove that there is only one possible group with each of these orders somehow, something to do with the group of order 1 being the trivial subgroup and finite cyclic groups of order $$p$$ having only one isomorphism class for a given $$p$$, I think.

2. Mr.Math

I know how to prove that a finite cyclic group of order $$p^2$$ has exactly $$3$$ subgroups. This follows directly from the Fundamental Theorem of Finite Abelian Groups. But I'm not sure how to show that it's the only group (up to isomorphism).

3. anonymous

I haven't actually even learned the FTowlet yet, I've been putting group theory on hold while doing some analysis work. To go the other direction with this proof though I think one would want to look at prime factorizations with 3 elements to show that it only works for $$p^2$$? Dunno.

4. anonymous

Haha, censored abbreviation, whoops.

5. Mr.Math

That sounds like a good idea.

6. anonymous

Intuitively, it is obvious that a number must be a squared prime to have a 3-element prime factorization. Rigorously, I'm not entirely sure how to approach showing that.

7. Mr.Math

Yep!

8. Mr.Math

I think I know how.

9. Mr.Math

Let's denote this group by $$G$$ where $$|G|=n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}$$. Then we know that $$G$$ and $$\{e\}$$ are subgroups. We want $$n$$ to be such that it has only one factor other than itself and $$1$$. This can only be if there exists a unique $$p_i|n$$ and $$p_i\ne n$$. So $$n=p_i^{\alpha_i}$$ for some $$1\le i \le r$$. Now if $$\alpha_i\ge 3$$, then $$p^3|n$$ and $$p^2|n$$, which contradicts our hypothesis. So $$\alpha_i=2$$.

10. anonymous

Sounds right to me. Thanks for reminding me that I need to get back to studying group theory, I've been putting it off for the last couple weeks :P

11. Mr.Math

You're welcome! And thanks for your help!