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mona123 October 12th, 2017 10:59 PM

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 $p^2\!$.


skipjack October 13th, 2017 12:57 AM

If the problem is referring to the subgroups of some group $G$, what are you told about $G$?

johng40 October 14th, 2017 08:55 AM

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

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.

SDK October 14th, 2017 09:29 AM

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 $p^2$.

johng40 October 15th, 2017 02:14 PM


I previously confessed that I can't solve problem 2. So I would be quite interested in seeing your solution. Please post it.

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