Abstract Algebra (Normal Subgroups)

In summary: H is in K?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\capK = {1} though.Any ideas about this one?Thanks!
  • #1
mattmns
1,128
6
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 [tex]\in[/tex]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[tex]\cap[/tex]K = {1}, prove that hk = kh for all h[tex]\in[/tex]H and k[tex]\in[/tex]K.
This one is really throwing me off.
We Know:
H and K are normal subgroups of a group G.
H[tex]\cap[/tex]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[tex]\cap[/tex]K = {1} though.
Any ideas about this one?

Thanks!
 
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  • #2
subgroup test comes to mind. oh wait never mind you have to prove normal subgroup sorry prolly not that easy then
 
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  • #3
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
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
 
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  • #5
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
for the 1st one you've got to show that the subgroup f(K) is normal in H, that is, [tex]f(K) = f(x)f(K)f(x)^{-1}[/tex] 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
mattmns said:
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}
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)
2) If H and K are normal subgroups of a group G satisfying H[itex]\cap[/itex]K = {1}, prove that hk = kh for all h in H and k in K.
This one is really throwing me off.
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
yeah i think i made a mistake. i should think harder
 
  • #9
Thanks everyone, I got them both :smile:
 

1. What is a normal subgroup?

A normal subgroup is a subgroup of a group that is invariant under conjugation by elements of the larger group. In other words, if an element of the larger group is conjugated with an element of the normal subgroup, the result will still be an element of the normal subgroup.

2. How is a normal subgroup different from a regular subgroup?

A regular subgroup is simply a subset of a group that satisfies the requirements for being a group itself. A normal subgroup, on the other hand, has the additional property of being invariant under conjugation by elements of the larger group.

3. What is the significance of normal subgroups in abstract algebra?

Normal subgroups play a crucial role in the study of group theory, as they allow for the creation of quotient groups. These quotient groups are useful in understanding the structure of larger groups, and can also be used to classify groups into different categories.

4. How can one determine if a subgroup is a normal subgroup?

There are a few different methods for determining if a subgroup is normal, including checking if the left and right cosets of the subgroup are equal, or verifying that the subgroup is the kernel of a homomorphism. In general, it is necessary to check both of these conditions for a subgroup to be considered normal.

5. Can a group have more than one normal subgroup?

Yes, a group can have multiple normal subgroups. In fact, every group has at least two normal subgroups: the trivial subgroup consisting of just the identity element, and the entire group itself. There can also be multiple normal subgroups of varying sizes within a single group.

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