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Math Help - uN = vN iff uv^-1 is in N??

  1. #1
    Senior Member Educated's Avatar
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    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)
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    Re: uN = vN iff uv^-1 is in N??

    Quote Originally Posted by Educated View Post
    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,
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    Re: uN = vN iff uv^-1 is in N??

    Quote Originally Posted by Plato View Post
    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...
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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.
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    Re: uN = vN iff uv^-1 is in N??

    Quote Originally Posted by johng View Post
    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!
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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, 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.
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  7. #7
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    Re: uN = vN iff uv^-1 is in N??

    Quote Originally Posted by Deveno View Post
    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.
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