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Subgroup Question

  1. May 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Let H be a subgroup of G.
    Prove Z(G) intersect H is contained in Z(H),
    use this result to verify that Z(G) intersect H is a normal subgroup of H.
    Given an example where Z(G) intersect H is a proper subgroup of Z(H).


    2. Relevant equations

    Z(G)= the center of G (x is an element of G |xa=ax for all a that is an element of G)

    3. The attempt at a solution
    a. wts Z(G) intersect H is contained in Z(H)
    So Z(G)=a is an element of G|ax=xa for all x that is an element of G. So Z(G) intersect H = a is an element of H|ax=xa for all x that is an element of G. Since an element a of Z(G) intersect H is in H and commutes with all of G, then it obviously commutes with all of H since H is a subgroup of G. Thus Z(G) intersect H is contained in H.

    b. LTS that Z(G) intersect H is a subgroup of H.
    Obviously Z(G) intersect H is contained within H as shown above.
    Since the identity e is always in Z(G) and in H (since H is a subgroup of G), then Z(G) intersect H is nonempty. Let a,b be elements of Z(G) intersect H. Then ab=ba by defintion of Z(G) intersect H. Let a(b^-1) be an element of Z(G) intersect H.
    Then a(b^-1)x=(b^-1)ax (since ab=ba implies (b^-1) = (a^-1)(b^-1)a.
    Thus, by the 1-step subgroup test, Z(G) intersect H is a subgroup of H.
    Since Z(G) intersect H = a is an element of H|ax=xa for all x that is an element of G,
    then Z(G) intersect H, then the element a obviously commutes with everything in H if it commutes with everything in G. Thus Z(G) intersect H is an abelain subgroup of H and is normal.
    3. My example is to consider D4 with H being <R90>. Then Z(G) intersect H= {R0, R180} which is a proper subgroup of Z(H) (Z(H)=<R90> which is of order 4).

    Is what I did proving all of this check or not. I am unsure about proving containments and subgroup tests. Thanks for any help
     
  2. jcsd
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