If K is a subgroup of G of order p^k, show that K is subgroup of H

[nowimpsbball]

Homework Statement

Let |G| = (p^n)m where p is prime and gcd(p,m) = 1. Suppose that H is a normal subgroup of G of order p^n. If K is a subgroup of G of order p^k, show that K is subgroup of H.

Homework Equations

The Attempt at a Solution

Okay, I wonder if there is more I need to do, or if I need to prove they are finite. I feel like I am missing something...but here is what I got
p^k has to be less than p^n because if p^k was bigger than p^n then p^k would not divide the order of G because p and m are relatively prime and K could not be a subgroup of G. The order of a subgroup must divide the order of the group.

Both H and K are subgroups of G, they both are closed under the same operation as G, and because n>k, p^k divides p^n and thus because K is closed under the operation of H and K's order divides the order of H, K must be a subgroup of H.

Thanks

 

[mighty2000]

for posting this question! it is important to carefully analyze and prove mathematical statements, as it allows for a deeper understanding and application of the concepts at hand. Let's break down the problem and see if we can find a solution.

First, let's recall some basic definitions and properties of groups. A group is a set G with a binary operation (typically denoted by *) that satisfies the following properties:
1. Closure: For any elements a and b in G, the result of the operation a*b is also in G.
2. Associativity: For any elements a, b, and c in G, (a*b)*c = a*(b*c).
3. Identity element: There exists an element e in G such that for any element a in G, e*a = a*e = a.
4. Inverse element: For any element a in G, there exists an element a^-1 in G such that a*a^-1 = a^-1*a = e.

Now, let's move on to the problem at hand. We are given that |G| = (p^n)m, where p is prime and gcd(p,m) = 1. We are also given that H is a normal subgroup of G of order p^n, and K is a subgroup of G of order p^k. We want to show that K is a subgroup of H.

To prove this, we need to show that K satisfies the four properties listed above. Since K is a subgroup of G, we know that it satisfies the first three properties. It is closed under the operation of G, it is associative, and it contains the identity element of G. So, we just need to show that it also contains the inverse element of each of its elements.

Let's take an element k in K. Since K is a subgroup of G, we know that k^-1 is also in K. Now, let's consider the element k*h, where h is an element in H. Since H is a normal subgroup of G, we know that k*h = h*k for all elements k in K and h in H. Therefore, (k*h)^-1 = h^-1*k^-1 = k^-1*h^-1. But we know that h^-1 is also in H, so k^-1*h^-1 is in K. This means that K contains the inverse element of k, and thus satisfies the fourth property.

Therefore,