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Thread: 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 $\displaystyle G$ be a group with operation $\displaystyle *$, and let $\displaystyle H$ be a subset of $\displaystyle G$. Then $\displaystyle H$ is a subgroup of $\displaystyle G$ iff

    (a) $\displaystyle H$ is nonempty

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

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

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

    Prove that $\displaystyle Z(G)$ 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 $\displaystyle G$ be a group with operation $\displaystyle *$, and let $\displaystyle H$ be a subset of $\displaystyle G$. Then $\displaystyle H$ is a subgroup of $\displaystyle G$ iff

    (a) $\displaystyle H$ is nonempty

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

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

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

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

    so let $\displaystyle a \in Z(G)$, then for all $\displaystyle x \in G$ we have $\displaystyle a*x = x*a$ ... then what. i need to show that $\displaystyle a^{-1}*x = x*a^{-1}$. 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 $\displaystyle a,b \in Z(G)$. Then for all $\displaystyle x \in G$, we have:

    $\displaystyle a * x = x * a$ ...............(1)

    and

    $\displaystyle b * x = x * b$ ...............(2)

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

    $\displaystyle a * x * b = x * a * b$

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

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

    thus, $\displaystyle a * b \in Z(G)$


    for (c):

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

    $\displaystyle a * x = x * a$

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

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

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

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

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

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

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

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

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

    This completes the proof



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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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