# Normal Subgroups proof.

1. Apr 17, 2012

### tiger4

1. The problem statement, all variables and given/known data

Let G be and group and let N and H be normal subgroups of G with \$N \subset H \subset G. Prove that H/N is a subgroup of G/N, and that it is a normal subgroup. Note that aKa^{-1} = {aKa^{-1} | k in K}.

2. Relevant equations

3. The attempt at a solution

I understand that since H is contained in G and N is contained in H that it would make sense that the factor group H/N is not only a subgroup, but a normal subgroup. However, I am struggling trying to figure out a way to transition from aKa^{-1} to either H or N. We've also learned the 2-step check of closure and inverses for proving subgroups, but i'm not quite sure how to apply that to factor groups. If I could get some suggestions where to start that would be great.

Last edited: Apr 17, 2012
2. Apr 17, 2012

### Chaos2009

Have you covered in your class that the elements of $H / N$ and $G / N$ are the cosets $N$ in $H$ and $G$? I might try this problem by taking an arbitrary element $h' \in H / N$ and $g' \in G / N$ and showing that $g' h' g'^{-1} \in H / N$. Since the elements chosen were arbitrary, $H / N \triangleleft G / N$.

3. Apr 19, 2012

### tiger4

let x,y be in aKa^-1.

thus x = aka^-1, for some k in K, and y = ak'a^-1 for some k' in K.

then xy^-1 = (aka^-1)(ak'a^-1)^-1 = (aka^-1)(ak'^-1a^-1) = a(kk'^-1)a^-1,

and since K is a subgroup kk'^-1 is in K whenever k,k' are, so xy^-1 is in aKa^-1.

thus aKa^-1 is certainly a subgroup of G.

what we need to do is show that aKa^-1 must be a subset of H.

note that K is a subgroup of H, hence aKa^-1 is a subset of aHa^-1. but H is normal in G, so aHa^-1 = H.

thus aKa^-1 is contained in H.

How does this look?