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- May 27th 2014, 02:58 PM #1

- May 27th 2014, 03:41 PM #2

- May 27th 2014, 04:42 PM #3
## Re: uN = vN iff uv^-1 is in N??

Sorry I don't think I stated the problem properly.

Let G be a group and N be a normal subgroup of G, ie . Then there is a theorem that states:

For we have if and only if .

Now my question is that if the group G is non-abelian, is it also true that: if and only if

So what you have there is the proof of the original theorem, which I understand (both directions). But the problem is that I want to show that which I can't seem to do because the inverse always seems to be on the left hand side...

- May 27th 2014, 06:54 PM #4

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## Re: uN = vN iff uv^-1 is in N??

Hi,

The operative word is "normal subgroup". If N is a normal subgroup of G and $u,v\in G$,

$uN=vN$ if and only if $uv^{-1}\in N$. The proof is almost immediate from the fact that N is a normal subgroup if and only if $uN=Nu$ for all $u\in G$. Now if you have trouble with the proof, post your problems.

If N is not a normal subgroup, your assertion is false. See if you can't find an example in $S_3$, where this is the full symmetric group on 3 letters. Again, if you have trouble, post your difficulties.

- May 28th 2014, 05:03 PM #5
## Re: uN = vN iff uv^-1 is in N??

- May 28th 2014, 07:17 PM #6

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## Re: uN = vN iff uv^-1 is in N??

Incidentally, the reason why we want normal subgroups is so that coset multiplication makes sense.

We would hope that (Na)(Nb) = N(ab).

If Ng = gN for all elements g of G, then a typical element of (Na)(Nb) = {nan'b: n,n' in N} is nan'b, and since aN = Na, an' = n''a, so that nan'b = (nn'')ab, which is in N(ab).

On the other hand we have for any element nab in N(ab): nab = an'b = (ea)(n'b), which is in (Na)(Nb).

If, however, Ng does not equal gN, for some particular g, the above argument doesn't work.

As johng has indicated, S_{3}provides an example: let N = {e, (1 2)}, and consider N(2 3) = {(2 3), (1 2 3)}, and N(1 3) = {(1 3), (1 3 2)}.

We would hope that N(2 3)N(1 3) = N((2 3)(1 3)) = N(1 2 3).

First, we calculate N(1 2 3) = {(1 2 3), (2 3)}.

Next, we calculate N(2 3)N(1 3):

(2 3)(1 3) = (1 2 3), so far, so good.

(2 3)(1 3 2) = (1 2) <---this is a problem, it isn't in N(1 2 3)

(1 2 3)(1 3) = (2 3), this would have been OK

(1 2 3)(1 3 2) = e, another problem.

As you can see, N(2 3)N(1 3) has 4 distinct elements, so it cannot possibly be ANY coset of N (all of which have just 2 elements).

So for that subgroup, there is no feasible way to define "coset multiplication".

***********

There is another way to look at this situation:

Suppose we define an equivalence relation ~ on a group G that respects the group multiplication: that is, if [g] is the equivalence class of g, whenever [g] = [g'] and [h] = [h'], we have [gh] = [g'h'].

We could then turn the set of equivalence classes into a group by setting: [g]*[h] = [gh].

I claim [e] is a subgroup of G. To see this, suppose a,b are in [e]. Then since [a] = [e] = [b], we have [ab^{-1}] = [a]*[b^{-1}] = [b]*[b^{-1}] = [bb^{-1}] = [e], so ab^{-1}is in [e].

Note this also show that if [a] = [b], that is if a ~ b, we have ab^{-1}in [e].

Finally, I claim that [e] is a NORMAL subgroup of G. For consider [ghg^{-1}], for h in [e]. We have [ghg^{-1}] = [g]*[h]*[g^{-1}] = [g]*[e]*[g^{-1}] = [geg^{-1}] = [gg^{-1}] = [e] = [h],

whence ghg^{-1}h^{-1}is in [e], so that ghg^{-1}is in h[e], which is a subset of [h]*[e] = [e]*[e] = [ee] = [e].

***************

There is also a third way of viewing this state of affairs:

Suppose we have a homomorphism f:G-->G'. We can view this as a surjective (onto) homomorphism f:G-->f(G).

Let K = {g in G: f(g) = e'}, where e' is the identity of G'. We can define an equivalence relation ~ on G by:

g ~ g' if and only if f(g) = f(g'), and we then have: K = [e]. We now show this equivalence relation respects the multiplication of G:

Suppose f(g) = f(g'), and f(h) = f(h'). Since f is a homomorphism, f(gh) = f(g)f(h) = f(g')f(h') = f(g'h'). As we saw above, this means K is a subgroup of G (it may be instructive to try to show this directly).

Next, we show that a ~ b if and only if Ka = Kb, that is, ab^{-1}is in K.

For suppose a ~ b. Then f(a) = f(b), so f(ab^{-1}) = f(a)f(b^{-1}) = f(b)f(b^{-1}) = f(bb^{-1}) = f(e) = e', so ab^{-1}is in K, so Ka = Kb.

On the other hand, suppose Ka = Kb, so that ab^{-1}is in K.

This means that f(ab^{-1}) = e', so f(a)f(b^{-1}) = e'. Multiplying both sides of this equation on the right by f(b), we get:

f(a)f(b^{-1})f(b) = f(b)

f(a)f(b^{-1}b) = f(b)

f(a)f(e) = f(b)

f(ae) = f(b)

f(a) = f(b), so a ~ b.

It is easy to show that K is normal (although from what we did above we don't need to):

Take gkg^{-1}, for any g in G, and k in K.

f(gkg^{-1}) = f(g)f(k)f(g^{-1}) = f(g)e'f(g^{-1}) = f(g)f(g^{-1}) = f(gg^{-1}) = f(e) = e', which show that gKg^{-1}is contained in K

(to see that K is contained in gKg^{-1}, write k = g(g^{-1}k'g)g^{-1}, and note that we have just shown g^{-1}kg is in K).

************

Now all of this may be a lot to take in, but I want to stress the inter-connectedness of 3 concepts:

1. Cosets of a NORMAL subgroup

2. Equivalences on a group that induce a group structure on the equivalence classes

3. Homomorphisms

these are like the top, side and front views of a three-dimensional object: they may appear to be quite different, but they refer to the same thing.

- May 29th 2014, 04:45 AM #7