Proving the Existence of Subgroups of Order pr in G of Order pem

[SNOOTCHIEBOOCHEE]

Homework Statement

Let G be a group of order p^em. Prove that G contains a subgroup of order p^r for every integer r $$\leq$$e

The Attempt at a Solution

By sylow 1 we know that there exists a subgroup of order p^e. so that case is done, we need to show that supgroups of order r<e exist now.

Thats tricky. I am guessing this is an application of the the thrid sylow theorem but i don't know how to use it.

s is the number of sylow p=subgroups. then s|m and is congruent to 1 mod p.

so we know that s has to be at least e. don't know where to go from here.

 

[mighty2000]

Thank you for your question. You are on the right track with using Sylow's theorems to prove the existence of subgroups of order pr in a group of order pem. Let me guide you through the steps of the proof:

1. By Sylow's third theorem, we know that the number of Sylow p-subgroups, denoted by s, divides the order of the group, which is pem. This means that s divides m.

2. We also know that s is congruent to 1 mod p, which means that s can be written as s = kp + 1 for some integer k.

3. Now, let's consider the group G acting on the set of all Sylow p-subgroups by conjugation. This action induces a permutation representation of G on the set of Sylow p-subgroups. By the orbit-stabilizer theorem, the size of each orbit divides the order of the group, which is pem.

4. Since the size of each orbit is a power of p, it follows that the size of each orbit must divide m. But we know that s, the number of Sylow p-subgroups, divides m. This means that each orbit must consist of exactly one Sylow p-subgroup.

5. Now, let H be the stabilizer of any Sylow p-subgroup P. By the orbit-stabilizer theorem, we know that |H| = |G|/|P| = pem/pe = pm. Since H is a subgroup of G, it follows that |H| divides |G|. Therefore, H is a subgroup of order pm, which is a power of p.

6. Finally, we can use the same argument as in step 4 to show that H must contain a subgroup of order pr for every integer r ≤ e. This is because the size of each orbit in H must divide |H|, which is pm, and since s divides m, each orbit must consist of exactly one subgroup of order pr.

Thus, we have shown that for any integer r ≤ e, G contains a subgroup of order pr. I hope this helps you understand the proof better. Keep up the good work in exploring group theory!