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

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

About quotient groups and coset representatives, there's this theorem that states $uN = vN \, \, \text{iff} \, \, u^{-1} v \in N$

I was wondering if you could switch the $u$ and $v^{-1}$ around to get something like:
$uN = vN \, \, \text{iff} \, \, u v^{-1} \in N$

(Assuming the group is non-abelian)

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

Originally Posted by Educated
About quotient groups and coset representatives, there's this theorem that states $uN = vN \, \, \text{iff} \, \, u^{-1} v \in N$
If $uN=vN$ then $\exists\{s,t\}\subset N$ such that $us=vt$. So
$us=vt\\st^{-1}=u^{-1}v\\st^{-1}\in N$.
That is one way. You post the other,

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

Originally Posted by Plato
If $uN=vN$ then $\exists\{s,t\}\subset N$ such that $us=vt$. So
$us=vt\\st^{-1}=u^{-1}v\\st^{-1}\in N$.
That is one way. You post the other,
Sorry I don't think I stated the problem properly.

Let G be a group and N be a normal subgroup of G, ie $N \unlhd G$. Then there is a theorem that states:
For $u, v \in G$ we have $uN = vN$ if and only if $v^{-1} u \in N$.

Now my question is that if the group G is non-abelian, is it also true that: $uN = vN$ if and only if $v u^{-1} \in N$

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 $v u^{-1} \in N$ which I can't seem to do because the inverse always seems to be on the left hand side...

4. ## 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.

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

Originally Posted by johng
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.
Ahh, yea I think I got it.

So the definition of a normal subgroup is $N \leq G$ such that $gNg^{-1} = N \, \, \, \forall \, \, \, g \in G$
Then just moving the $g^{-1}$ to the other side, we get $N \leq G$ such that $gN = Ng \, \, \, \forall \, \, \, g \in G$

Then it's just a simple case of $uN = Nu = Nv = vN$ and dealing with the same proof for the right cosets to get $uv^{-1}\in N$

Thanks!

6. ## 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, S3 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-1h-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-1b) = 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-1k'g)g-1, and note that we have just shown g-1kg 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.

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

Originally Posted by Deveno
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, S3 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".
Never thought of it that much in depth, but I see why they defined a 'normal' subgroup the way they do now. It's all just so coset multiplication works.