Homework Help: Abstract Algebra (Normal Subgroups)

1. Nov 1, 2005

mattmns

Hello. We got a review today in abstract algebra, and I am stuck on two problems.

1) Let f: G -> H be a surjective homomorphism of groups. Prove that if K is a normal subgroup of G, then f(K) is a normal subgroup of H. Where f(K)= {f(k): k $$\in$$K}
The entire f(K) part is really throwing me off.
We know:
G and H are groups
K is a normal subgroup of G
Also H is a normal subgroup because im_f = H (because f is surjective)
The real problem is I don't quite understand the whole f(K) part. Any ideas about what f(K) is, and any ideas about the problem?

2) If H and K are normal subgroups of a group G satisfying H$$\cap$$K = {1}, prove that hk = kh for all h$$\in$$H and k$$\in$$K.
This one is really throwing me off.
We Know:
H and K are normal subgroups of a group G.
H$$\cap$$K = {1} is a normal subgroup of G. (because of a previous problem I had proved).
Also, HK is a subgroup (by a theorem from my book)
For this problem it seems that it would be sufficient to show that G is abelian, but I am not sure how we would do that. Or maybe just use some general properties algebraically to show that hk=kh. I am not sure how we are supposed to use the H$$\cap$$K = {1} though.

Thanks!!!

Last edited: Nov 1, 2005
2. Nov 1, 2005

1800bigk

subgroup test comes to mind. oh wait never mind you have to prove normal subgroup sorry prolly not that easy then

Last edited: Nov 1, 2005
3. Nov 1, 2005

mattmns

Yes, but I am not understanding how I would say that 1 is in f(K), or x,y in f(K) => xy in f(K), my real problem is understanding how I would use f(K) to see if the properties of a subgroup apply to f(K). Any ideas about f(K)?

4. Nov 1, 2005

1800bigk

yes I had a problem like this but it was dealing with isomorphisms I think I assumed f(a) and f(b) were in f(k) then I said f(a)f(b) was in f(k) since ab is in k so f(ab) is in f(k) and f(ab)=f(a)f(b), similar argument holds for a inverse

f(k) is non empty because f(e) is in f(k)

since k is a group it has all the properties we need like identity and a,b in k implies ab in k so f(ab) is in f(k)
a in k then a^-1 in k so f(a^-1) is f(k)

f(a^-1) = (f(a))^-1

Last edited: Nov 1, 2005
5. Nov 1, 2005

mattmns

Hmm, that is an interesting idea, I think I remember my teacher doing something similar, I will try it out, thanks. Any ideas for #2?

6. Nov 2, 2005

fourier jr

for the 1st one you've got to show that the subgroup f(K) is normal in H, that is, $$f(K) = f(x)f(K)f(x)^{-1}$$ for any f(x) in H if K is normal in G.
unless i made a mistake (or i'm getting better at algebra) it's not very hard. all you've got to use, after all, is properties of homomorphisms to show that the image is a normal subgroup. i think the reason you need a surjective homomorphism is so that you get a group in the image, otherwise you could get the empty set as an image which isn't a normal subgroup.
for the 2nd one i'd try out some examples. the quaternions is a noncommutative group in which every subgroup is normal, so you're idea won't work (showing it's abelian). try playing with a concrete example like the quaternions & see if that helps.

7. Nov 2, 2005

AKG

Suppose x is in f(K) and h is in H. To show f(K) is normal, you want to show that hxh-1 is in f(K). Well you know that since f is surjective, there is some g in G such that f(g) = h. Since x is in f(K), you know there is some k in K such that f(k) = x. So you can pick g and k such that:

hxh-1 = ... (figure the rest out)
Do you know that if H is normal, then for every g in G, gH = Hg? In other words, for every g in G and for every h in H, there is some h' in H (possibly the same as h) such that gh = h'g. This holds for every g in G, including every g in K. Can you go from here?

8. Nov 2, 2005

fourier jr

yeah i think i made a mistake. i should think harder

9. Nov 2, 2005

mattmns

Thanks everyone, I got them both