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Homework Help: Help showing that G/H is a group?

  1. May 14, 2010 #1
    Hi guys

    I was just wondering whether anyone could help me with Group Theory

    I am trying to prove that G/H is a group (where G is a group and H is a normal subgroup of G)

    I know that i need to go through the 4 properties, Identity, inverse, Associativity and Closure but i'm not sure where to start. I'm also not sure how to properly define the composition Law in G/H

    Is anybody able to help?
  2. jcsd
  3. May 14, 2010 #2
    The elements of G/H are cosets of the form aH, where a is in G. If aH and bH are in G/H, the composition law is given by (aH)(bH) = (ab)H. In addition to whatever group properties, you must also show that this law of composition is well-defined. That is, if aH = a'H and bH = b'H, then (aH)(bH) = (a'H)(b'H).
  4. May 14, 2010 #3


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    First start by defining the group operation.

    The elements of G/H are the cosets of H. (Let's say left cosets for definiteness. It actually doesn't matter because left cosets and right cosets are the same thing for normal subgroups, but I don't know if you have proved that yet.)

    Thus [tex]G/H = \{gH : g \in G\}[/tex].

    I claim that the following is a well-defined operation on G/H:

    [tex](aH)(bH) = (ab)H[/tex]

    However, before you can proceed, you need to verify that this is well-defined. Why is this an issue? It's because the coset [itex]aH[/itex] has multiple elements (assuming H is not trivial), and [itex]a[/itex] is only one of the elements. Therefore you must show that if [itex]a_1 H = a_2 H[/itex] and [itex]b_1 H = b_2 H[/itex], then [itex](a_1 b_1)H = (a_2 b_2)H[/itex].

    Once you have done that, you can proceed with verifying that the group axioms hold. (Hint: H is the identity element.)
  5. May 14, 2010 #4
    Thanks for the help, i was just wondering how i go about showing that it was well defined. I tried to use the facts that were given

    I didn't know what to do once (aH)(bH) = (ab)H....is the next step to say that this is ab'H?
  6. May 14, 2010 #5
    I have now come to this conclusion:

    i need to show that (ab)H = (a'b')H

    First step: (ab)H = aHbH = a'Hb'H = (a'b')H

    Is that how to show its well defined or is there anything that i wasnt allowed to do?
  7. May 14, 2010 #6


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    No, if the goal is to show that an operation is well-defined, you can't use that operation (thereby implicitly assuming it's well-defined) as part of the proof. Also, you didn't use the fact that H is a normal subgroup of G, which should be a red flag that something is wrong.

    Try proceeding this way instead:

    We assume that aH = a'H and bH = b'H.


    abH = ab'H

    Now, *if* it were true that b'H = Hb', then we could continue this way:

    abH = ab'H = aHb' = a'Hb' = a'b'H

    and we would be done.

    So all we need to do is show that b'H = Hb'. How do we do that? Hint: we still haven't used the fact that H is normal.
  8. May 14, 2010 #7
    For a group to be normal subgroup of another doesnt it mean gHg-1 belongs to H? I don't see how this would help?
  9. May 14, 2010 #8


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    OK, so your definition is that H is normal in G if

    [tex]gHg^{-1} \subset H[/tex]

    for every [itex]g \in G[/itex], right?

    I claim that something stronger is true:

    [tex]gHg^{-1} = H[/tex]

    for every [itex]g \in G[/itex]. Assuming my claim is true, then I can simply multiply both sides on the right by [itex]g[/itex] to get

    [tex]gH = Hg[/tex]

    (Here I used [itex]g^{-1}g = 1[/itex].)

    Now substitute [itex]b'[/itex] for [itex]g[/itex] and you get the desired

    [tex]b'H = Hb'[/tex]

    So now you just need to prove my claim. I suggest starting with

    [tex]gHg^{-1} \subset H[/tex]

    and think about how you can morph that into

    [tex]H \subset g^{-1}Hg[/tex]

    for all [itex]g \in G[/itex].
  10. May 15, 2010 #9
    Can we just multiply both sides on the left by g-1 and the right by g so that it now becomes the required result? i.e. H is a subset of g-1Hg?
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