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Mr.Math

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 \(p^2\).

  • one year ago
  • one year ago

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  1. nbouscal
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    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.

    • one year ago
  2. Mr.Math
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    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).

    • one year ago
  3. nbouscal
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    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.

    • one year ago
  4. nbouscal
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    Haha, censored abbreviation, whoops.

    • one year ago
  5. Mr.Math
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    That sounds like a good idea.

    • one year ago
  6. nbouscal
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    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.

    • one year ago
  7. Mr.Math
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    Yep!

    • one year ago
  8. Mr.Math
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    I think I know how.

    • one year ago
  9. Mr.Math
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    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 \(\alpha_i=2\).

    • one year ago
  10. nbouscal
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    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

    • one year ago
  11. Mr.Math
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    You're welcome! And thanks for your help!

    • one year ago
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