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 October 12th, 2017, 11:59 PM #1 Senior Member   Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 Number of subgroups Please help me to prove the following problem: >1) Every subgroup of order $p^2$ contains $1\;(\text{mod }p)$ different subgroups of order $p$. >2) Every subgroup of order $p$ is contained in $1\;(\text{mod }p)$ different subgroups of order $p^2\!$. Thanks. Last edited by skipjack; October 13th, 2017 at 01:36 AM. October 13th, 2017, 01:57 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,127 Thanks: 2336 If the problem is referring to the subgroups of some group $G$, what are you told about $G$? October 14th, 2017, 09:55 AM #3 Member   Joined: Jan 2016 From: Athens, OH Posts: 93 Thanks: 48 Your last couple of posts are related to a theorem of Helmut Wielandt. If G is a finite group of order $p^kn$ where p is a prime, then the number of subgroups of G of order $p^k$ is congruent to 1 mod p. Here, it is not assumed that p is prime to n. A proof can be found here https://math.stackexchange.com/quest...sylows-theorem Now as to your specific questions (I assume p is prime and $p^2$ divides the order of G.) 1. Let H be a subgroup of order $p^2$. Either H is cyclic or H is the direct product of 2 subgroups of order p. In the first case, H has exactly one subgroup of order p. In the second case, H has $p^2-1$ elements of order p and hence ${p^2-1\over p-1}=p+1$ subgroups of order p. 2. I can't prove this without Wielandt's theorem. However, it is true and here's a proof. Let H be a subgroup of order p and $S=\{K:H\subset K\text{ and K is a subgroup of order }p^2\}$. Let L be the subgroup generated by all the elements of S (S is not empty). Since each $K\in S$ is abelian, H is a normal subgroup of L. By Wielandt, the number of subgroups of L/H of order p is 1 mod p. That is, the cardinality of S is 1 mod p. October 14th, 2017, 10:29 AM #4 Senior Member   Joined: Sep 2016 From: USA Posts: 685 Thanks: 461 Math Focus: Dynamical systems, analytic function theory, numerics Here are some hints: Start by proving that any group of order $p^2$ is abelian. To prove this consider such a group mod its center. What can you say if this quotient group is cyclic? Now apply Sylow's theorem to any group of order $p^2$. For 2, I would suggest looking at equivalence classes of an element of order $p$ under conjugation. Apply the orbit stabilizer theorem to count the ways an element of order $p$ can belong to a subgroup of order $p^2$. Last edited by skipjack; October 14th, 2017 at 04:27 PM. October 15th, 2017, 03:14 PM #5 Member   Joined: Jan 2016 From: Athens, OH Posts: 93 Thanks: 48 SDK, I previously confessed that I can't solve problem 2. So I would be quite interested in seeing your solution. Please post it. Tags number, subgroup, subgroups Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post thippli Abstract Algebra 2 May 27th, 2013 09:26 AM gaussrelatz Algebra 1 October 11th, 2012 12:30 AM honzik Abstract Algebra 0 February 13th, 2011 07:45 AM DanielThrice Abstract Algebra 1 November 25th, 2010 03:28 PM bjh5138 Abstract Algebra 2 August 14th, 2008 01:03 PM

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