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 $p^2$.

Question

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 $p^2$.

anonymous · Answer

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.

mr.math · Answer

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).

anonymous · Answer

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.

anonymous · Answer

Haha, censored abbreviation, whoops.

mr.math · Answer

That sounds like a good idea.

anonymous · Answer

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.

mr.math · Answer

Yep!

mr.math · Answer

I think I know how.

mr.math · Answer

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 $\alpha_i=2$.

anonymous · Answer

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

mr.math · Answer

You're welcome! And thanks for your help!