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Proving a subgroup

  1. Oct 15, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    I got this question from contemporary abstract algebra :

    http://gyazo.com/7a9e3f0603d1c0dcfde256e7b05276cd

    2. Relevant equations

    One step subgroup test :
    1. Find my defining property.
    2. Show that my potential subgroup is non-empty.
    3. Assume that we have some a and b in our potential subgroup.
    4. Prove that ab-1 is in our potential subgroup.

    3. The attempt at a solution

    1. Defining property : xh = hx for x in G and for all h in H.
    2. C(H) ≠ ∅ because the identity element e is in C(H) and satisfies xe = ex.
    3. Suppose a and b are in C(H), then xa = ax and xb = bx.

    4. Show that ab-1 is in H whenever a and b are in H. So we want : xab-1 = ab-1x

    Start with :

    xa = ax
    x(ab-1) = (ax)b-1
    x(ab-1) = (xa)b-1
    x(ab-1) = x(ab-1)
    x(ab-1) = (ab-1)x

    I know this is probably horribly wrong, but for some reason I can't seem to see how to do this properly. Any help would be appreciated.
     
  2. jcsd
  3. Oct 15, 2012 #2

    Dick

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    You skipped a step. You need to show that if bx=xb then b^(-1)x=xb^(-1). You can't just assume it.
     
  4. Oct 15, 2012 #3

    Zondrina

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    Ahhhh so it would be a sort of two step thing here.

    First we want to show xb-1 = b-1x. So :

    xb = bx
    b-1xbb-1 = b-1bxb-1
    b-1xe = exb-1
    b-1x = xb-1

    Now we can show that xab-1 = ab-1x. So :

    xa = ax
    x(ab-1) = a(xb-1)
    x(ab-1) = (ab-1)x
     
  5. Oct 15, 2012 #4

    Dick

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    Yes, that's it.
     
  6. Oct 15, 2012 #5

    Zondrina

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    So we could conclude that C(H) ≤ G as desired. Thanks a bundle for your help :)
     
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