# Factor Groups

1. Nov 11, 2009

### lmedin02

1. The problem statement, all variables and given/known data
Let G be a group, and let H be a normal subgroup of G. Must show that every subgroup K' of the factor group G/H has the form K'=K/H, where K is a subgroup of G that contains H.

2. Relevant equations
I dont see how to get started.

3. The attempt at a solution
I wrote down the definitions of each factor group, G/H and K/H. If K'=K/H, then H must be normal in K.

2. Nov 25, 2009

### lmedin02

To prove this fact I tried to show that K' is contain in K/H for some subgroup H in G and vice versa.

3. Nov 25, 2009

### breedencm

It might help thinking of this problem with the natural homomorphis, g: G -> G/H. From this you can show that for any subgroup of K' of G/H, g^-1(K') is a subgroup of G, containing K. Then maybe an isomorphism theorem jumps at you after that?

Note: g^-1(K') is the pre-image of g, ie; g^-1(K') = { a in G | g(a) is in K' }

4. Nov 26, 2009

### lmedin02

I created an example where this theorem holds and observe that K must be defined as the set containing all elements in each element of K' (i.e., in each coset of G/H that is contain in K'). With this choice of K, the theorem works. I am now having alittle trouble justifying that K is indeed a subgroup.

5. Nov 26, 2009

### breedencm

Well, this was what I was trying to say:

Let H be a normal subgroup of G. Let g: G -> G/H. Let K' be a subgroup of G/K. We claim that g^-1(K') = { a in G | g(a) is in K' } forms a subgroup in G, which contains H (and actually we claim to show, later that this is the subgroup we wish to find).

K = g^-1(K') is a subgroup of G:

K is not empty, and contains H: In fact, H (which is not empty) is a subset of G, since for any h in H, g(a) = hH = eH (the identity element in G/H, which is clearly an element of K' since K' is a group).

K is closed: let a, b be in K, then g(a) is in K', and g(b) is in K', so g(ab) = g(a)g(b) is in K' since K' is a group.

K contains inverses: let a be in K, then g(a) is in K', so g(a)^-1 = g(a^-1) is in K', so a^-1 is in K.

Ok, so now we show there is an isomorphism between K/H and K'. But this should be clear to you since H is the kernel of the map of g.

So as we've shown, for each K' a subgroup of G/H is isomorphic to a subgroup K/H, K a subgroup of G.

6. Nov 29, 2009

### lmedin02

I got it, thank you. I did not consider defining this mapping. It follows by closure that g is a homomorphism and that the Kernel of g is normal to K. Then by the first isomorphism theorem it follows that the factor group K/(ker g) is isomorphic to K'.