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?