abstract algebra help: commutator subgroups

Let G be a finite non-abelian group with no elements of order 2, and let H be a normal subgroup containing the commutator subgroup of G. Show that the product of all elements of G (written in any order) lies in H.

I am really lost! I would start by stating that: Since H is normal and contains the commutator subgroup of G then G/H is a commutative group. So abH = aH bH = bH aH = baH.

but I don't really know where to go from here

Re: abstract algebra help: commutator subgroups

I think you're almost there. In a finite abelian group G, let x be the product of all elements of G. Prove . (If you can't do this, post again for help.) Then if G has no elements of order 2, x = 1. Now compute in G/H.

Re: abstract algebra help: commutator subgroups

but G is supposed to be a finite-non abelain group

i tried to finish the proof I added this, do you think this is complete or am I still missing something?

You can pair up every g in G with it's inverse g^-1, and since G has no elements of order 2 then g <> g^-1. So the number of elements in G is of the form 2r + 1, where r is the number of pairs (g,g^-1) and 1 is for the identity.

Write P = product of all elements in G (in any arbitrary order). Then PH = product of the gH across all g in G in the order given by P. Since G/H is abelian then we can reorder the product of the gH's so that they cancel out with their inverses, ie, move each gH next to g^-1H then multiply to get (gg^-1)H = H. In so doing we have PH = H, which means that P is in H, and we're done.

Re: abstract algebra help: commutator subgroups

do u think this is complete or would you explain it differently?

Re: abstract algebra help: commutator subgroups

Hi,

Yes, it looks like a valid proof to me. In fact, easier than what I had in mind in my previous post.