Proving Normality of Subgroups in Factor Groups: A Step-by-Step Approach

[tiger4]

Homework Statement

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}.

Homework Equations

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.

 

[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.

 

[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.


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