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Math Help - Abstract Algebra: Subgroup Proof

  1. #1
    is up to his old tricks again! Jhevon's Avatar
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    Abstract Algebra: Subgroup Proof

    Hail Mathematicians,

    It's me again.

    I have a problem, here is the appropriate theorem:

    Theorem: Let G be a group with operation *, and let H be a subset of G. Then H is a subgroup of G iff

    (a) H is nonempty

    (b) if a \in H and b \in H, then a*b \in H, and

    (c) if a \in H, then a^{-1} \in H

    i've used this theorem over and over in studying, but i have trouble proving (b) and (c) in the following problem


    Problem:

    Assume that G is a group with operation * and let Z(G) = \{ x \in G : a*x = x*a \mbox{ for every } a \in G \}

    Prove that Z(G) is a subgroup of G.



    This is a homework problem, so only hints please.

    Thanks, guys and gals

    EDIT: Proving it is nonempty is trivial, so don't even bother...
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  2. #2
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    Quote Originally Posted by Jhevon View Post
    Hail Mathematicians


    I have a problem, here is the appropriate theorem:

    Theorem: Let G be a group with operation *, and let H be a subset of G. Then H is a subgroup of G iff

    (a) H is nonempty

    (b) if a \in H and b \in H, then a*b \in H, and

    (c) if a \in H, then a^{-1} \in H

    i've used this theorem over and over in studying, but i have trouble proving (b) and (c) in the following problem[B]
    I never used that theorem. I simply do it by definition: (i) H is a non-trivial subset of G (ii) if a*b are in H for all a,b in H (iii) e is in H (iv) if a in H then a^(-1) in H (v) * is associate (but that follows because G is a group).

    Try doing it this way.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post



    I never used that theorem. I simply do it by definition: (i) H is a non-trivial subset of G (ii) if a*b are in H for all a,b in H (iii) e is in H (iv) if a in H then a^(-1) in H (v) * is associate (but that follows because G is a group).

    Try doing it this way.
    yes. (i) and (iii) are easy. but i can't show that a*b is in Z(G) for all a,b in Z(G).

    so far i have: Let a,b \in Z(G). Then for all x \in G we have a*x = x*a and b*x = x*b ...and there i'm stuck. i need to show that a*b \in Z(G)

    also, i need to show that for all a \in Z(G) we have a^{-1} \in Z(G), which is what i think you are saying in (iv)

    so let a \in Z(G), then for all x \in G we have a*x = x*a ... then what. i need to show that a^{-1}*x = x*a^{-1}. how do i do that?
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    is up to his old tricks again! Jhevon's Avatar
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    does this work?

    for (b):

    Let a,b \in Z(G). Then for all x \in G, we have:

    a * x = x * a ...............(1)

    and

    b * x = x * b ...............(2)

    Take equation (1) and apply b to the both sides, we get:

    a * x * b = x * a * b

    \Leftrightarrow a * b * x = x * a * b ..........by equation (2)

    \Leftrightarrow (a * b) * x = x * (a * b)

    thus, a * b \in Z(G)


    for (c):

    Let a \in Z(G), and e \in G be the identity element of G. Then for all x \in G we have:

    a * x = x * a

    apply a^{-1} to both sides. (we know a^{-1} exists since a \in G, and G is a group, so a^{-1} \in G).

    \Rightarrow a^{-1} * a * x = a^{-1} * x * a

    \Rightarrow e * x = a^{-1} * x * a

    \Rightarrow x = a^{-1} * x * a

    apply a^{-1} again, this time to the right most position

    \Rightarrow x * a^{-1} = a^{-1} * x * a * a^{-1}

    \Rightarrow x * a^{-1} = a^{-1} * x * e

    \Rightarrow x * a^{-1} = a^{-1} * x

    thus, a^{-1} \in Z(G)

    This completes the proof



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  5. #5
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    By the way the subgroup Z(G) has a name, it is called the "center" of the group.

    Let x and y be in the center. We ask whether xy is in the center, if and only if, for any g in G we have g(xy) = (xy)g. But that is true because g(xy) = (gx)y = (xg)y=x(gy) = (xy)g.
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