I am having trouble beginning and understanding how to tackle this problem. I've seen examples in class in which we use group Zn and Sn but I don't know how to work with this particular example. Am I making it too hard?

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- November 26th 2012, 06:10 PMbondvalencebondNormal Subgroups using The First Isomorphism Theorem
I am having trouble beginning and understanding how to tackle this problem. I've seen examples in class in which we use group Zn and Sn but I don't know how to work with this particular example. Am I making it too hard?

Attachment 25946 - November 26th 2012, 09:08 PMbondvalencebondRe: Normal Subgroups using The First Isomorphism Theorem
Just in case anyone is interested, this is what I've worked up so far:

There is a canonical isomorphism

(G/A)/(B/A) ≅ G/B

If you can construct a homomorphism φ:G/A → G/B, show that it is well-defined, then show that its kernel is B/A, you're done with all parts of the problem.

For B, this amounts to showing φ does not depend on choice of coset representatives. Assume you have x and y such that φ(xA) ≠ φ(yA). Prove that xA ≠ yA.

For C, just work straight from the definition of homomorphism. Specifically, set it up by choosing arbitrary x, y ∈ G, then show φ(xA yA) = φ(xA)φ(yA). It's pretty much immediate.

For D, suppose you have b ∈ B. Where does φ(bA) map? φ(bA) = bB = eB which is the identity, so bA ∈ ker φ. This proves A as well since the kernel of a homomorphism is always a normal subgroup.

The canonical isomorphism listed above is the conclusion of part E. - November 27th 2012, 09:24 AMDevenoRe: Normal Subgroups using The First Isomorphism Theorem
first things first:

we need to show that B/A is a normal subgroup of G/A. it's not hard to show it's a subgroup:

suppose bA,b'A are two (left) cosets of A with b,b' in B (these cosets exist, since any coset gA exists as an element of G/A for any element g of G).

now (bA)(b'A)^{-1}= (bA)(b'^{-1}A) = (bb'^{-1})A (considering bA and b'A as elements of G/A).

so to prove B/A is a subgroup of G/A, all we need to do is show is that (bA)(b'A)^{-1}is in B/A (the "one-step test").

and since (bA)(b'A)^{-1}= (bb'^{-1})A, it suffices to show bb'^{-1}is in B, which it is since B is a subgroup.

well is it a NORMAL subgroup? what would that mean? it would mean:

(gA)(bA)(gA)^{-1}is in B/A, for any gA in G/A. so what is (gA)(bA)(gA)^{-1}?

(gA)(bA)(gA)^{-1}= (gbg^{-1})A (by the defintion of coset multiplication in G/A). and to show THAT coset is in B/A,

all we need to do is show gbg^{-1}is in B. but B is normal in G, so.....

so that's (A).

now for (B): well the "obvious" thing to try to do is define:

φ(gA) = gB.

but we need to be sure that we always get the same result gB, now matter which element of gA we use to represent gA.

in other words, if gA = g'A, we need to have gB = g'B, or else φ won't be a function. (it has to only depend on the coset gA, not the element g).

so, if gA = g'A, this mean g'^{-1}g is in A (because g'_{-1}(gA) = g'^{-1}(gA) = (g'^{-1}g)A = eA = A, and since e is an element of A,

g'^{-1}(ge) = g^{-1}g is thus an element of A).

but A is a subgroup of B, so if g'^{-1}g is in A, it's AUTOMATICALLY in B, so gB = g'B. so φ is well-defined.

(C) is, as you remarked, pretty straight-forward. coset multiplication works the same way in G/A as in G/B.

a word about (D): showing that every coset bA is in ker(φ) only shows that ker(φ) contains B/A. it might be even bigger.

so what you still have to do is show that if gA is in ker(φ), that g is in B.

but what does it mean for gA to be in ker(φ)? it means φ(gA) = B (since B is the identity of G/B).

since φ(gA) = gB, that means gB = B, which means that g is in B.

it IS important that φ be onto, or else we cannot derive the isomorphism. however, given gB in G/B, it's pretty clear that gA is a pre-image. - November 28th 2012, 11:10 AMbondvalencebondRe: Normal Subgroups using The First Isomorphism Theorem
Do you mind explaining part D a bit more? I think I'm missing something...

- November 28th 2012, 11:53 AMbondvalencebondRe: Normal Subgroups using The First Isomorphism Theorem
Apparently for part D, it is not enough to show that... Any ideas on how to clarify it?

This is what I wrote:

If gA is in kerφ, then g is in B. Since φ(gA) = gB and gB=B then g is in B. Suppose b is in B then φ(bA) = bB = eB therefore bA is in kerφ.

My professor stated that she didn't think it was enough. - November 28th 2012, 12:31 PMjakncokeRe: Normal Subgroups using The First Isomorphism Theorem
Maybe she was looking for an if and only if statement (A stronger statement), which this is

- November 28th 2012, 03:18 PMbondvalencebondRe: Normal Subgroups using The First Isomorphism Theorem
I saw your suggestion after I turned my work in but I did include that. Thanks!

- November 28th 2012, 03:21 PMDevenoRe: Normal Subgroups using The First Isomorphism Theorem
if you want to show that ker(φ) = B/A, it's not enough to show containment "one-way". also, the order of the statements you make matters.

if gA is in ker(φ), then φ(gA) = B, so gB = B. you can't just jump to what you want to prove.

take a typical element gb in gB. since gB = B, we have gb = b', for some other element b' of B. so g = b^{-1}b', which is in B, since B is a subgroup.

this shows that ker(φ) is contained in {gA in G/A: gA = bA for some b in B} = {bA: b in B} = B/A.

you should put the part starting with "since" before the part saying "then g is in B".

aside from that small grammatical snafu, you have shown that ker(φ) and B/A (as a subset of G/A) contain each other, which means they are equal sets.

in general, when you want to show ker(f) = K, for some subset K of a group G, (where f:G-->G' is some homomorphism) you have to do two things:

1. show that f(k) = e', for every k in K. this shows the kernel contains K.

2. show that if something is in the kernel, it MUST lie within K. this shows K contains the kernel.

an alternative to #2, is to show that if g is NOT in K, then f(g) is NOT the identity of g' (this is the contrapositive of #2), so g is not in the kernel.

for isomorphisms (NOT homomoprisms) it suffices to show #2 (because #1 is trivial for the subgroup {e} of G).