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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 \(p^2\).
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 \(p^2\).

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 4 years ago
Best ResponseYou've already chosen the best response.1I 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
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0Haha, censored abbreviation, whoops.

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.1That sounds like a good idea.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Intuitively, it is obvious that a number must be a squared prime to have a 3element prime factorization. Rigorously, I'm not entirely sure how to approach showing that.

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.1Let'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 \(p_in\) 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 \(p^3n\) and \(p^2n\), which contradicts our hypothesis. So \(\alpha_i=2\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sounds 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
 4 years ago
Best ResponseYou've already chosen the best response.1You're welcome! And thanks for your help!
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