Prove Normal Subgroup When G Order is Twice H Order

[Nevermore]

I'm working on introductory group theory and am stuck on this proof. I don't even know where to start, so I'd appreciate any help at all!

"A subgroup H of a finite group G is said to be a normal subgroup if, for each element h ∈ H and each element g ∈ G, the element g^1hg ∈ H

Prove that if the order of G is twice the order of H, then H is a normal subgroup of G."

 

[matt grime]

Take the definition of normal, and play with it. What it says is that for all g in G, the set gHg^{-1}=H, or that gH=Hg, ie left and right cosets agree. Suppose H has index two (i.e. |G|=2|H|). How many left cosets are there? How many right cosets are there. What properties of cosets do you know? How about cosets are disjoint or equal? Can you see how this implies the result?

 

[Architeuthis Dux]

To prove this statement, we need to show that for every element h ∈ H and every element g ∈ G, the element g^1hg ∈ H.

Let's first define the order of a group as the number of elements in the group. So if G has order n, then it has n elements. Similarly, if H has order m, then it has m elements.

Now, since we know that the order of G is twice the order of H, we can write this as n = 2m. This means that for every element h ∈ H, there must exist two elements g ∈ G such that g^1hg ∈ H. This is because for every element in H, there must be at least two elements in G that can be paired with it to satisfy the condition of being in a normal subgroup.

To further prove this, let's consider the left cosets of H in G. A left coset of H in G is a set of elements in G that are obtained by multiplying each element of H by a fixed element g ∈ G. So for a given g ∈ G, the left coset of H in G is given by the set gH = {gh | h ∈ H}.

Since we know that the order of G is twice the order of H, we can divide G into two equal parts, each containing m elements. Let's call these two parts A and B. Now, for any element g ∈ A, the left coset gH will contain m elements (since H has m elements). Similarly, for any element g ∈ B, the left coset gH will also contain m elements. This means that for every element h ∈ H, there will be exactly two elements g ∈ G (one in A and one in B) that will result in g^1hg ∈ H.

Therefore, we can conclude that H is a normal subgroup of G, since for every element h ∈ H and every element g ∈ G, the element g^1hg ∈ H. This proves that if the order of G is twice the order of H, then H is a normal subgroup of G.