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Math Help - if ab=ca, how to get b=c???

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    if ab=ca, how to get b=c???

    Let a, b, c \in G where G is a finite group.
    If ab=ca, what conditions we need in order to have b=c???
    conjugacy class?? order??
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    Re: if ab=ca, how to get b=c???

    Hey deniselim17.

    Think about applying various post and pre multiplications of b inverse and c inverse. For example if we apply b inverse post multiplication on LHS we get a = cab^(-1) which is also equal to c^(-1)ab.

    You can also get the identity in terms of a,b,c and b,c inverses.

    This should give you a starting point to answering your question (you can use these identities to infer properties of a given that b = c).
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    Re: if ab=ca, how to get b=c???

    Given ab=ca
    if b=c, ba=ca=ab
    The condition for b=c is that G be commutative
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    Re: if ab=ca, how to get b=c???

    Quote Originally Posted by Hartlw View Post
    The condition for b=c is that G be commutative
    And that  a \ne 0.
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    Re: if ab=ca, how to get b=c???

    Quote Originally Posted by ebaines View Post
    And that  a \ne 0.
    Since the OP is dealing with groups, every element has an inverse. Hence, 0 is not an element of any multiplicative group, and there is no need to ensure a \neq 0. Hartlw is correct that the condition deals with commutativity. But, it is not necessary that the entire group be commutative. Instead, it is only necessary that a \in Z(G) where Z(G) = \{z \in G \mid \forall g\in G, zg = gz\} is the center of the group.
    Thanks from Deveno
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    Re: if ab=ca, how to get b=c???

    Quote Originally Posted by SlipEternal View Post
    Since the OP is dealing with groups, every element has an inverse. Hence, 0 is not an element of any multiplicative group, and there is no need to ensure $$a \neq 0$$. Hartlw is correct that the condition deals with commutativity. But, it is not necessary that the entire group be commutative. Instead, it is only necessary that $$a \in Z(G)$$ where $$Z(G) = \{z \in G \mid \forall g\in G, zg = gz\}$$ is the center of the group.
    redone for new LaTex
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    Re: if ab=ca, how to get b=c???

    Quote Originally Posted by Hartlw View Post
    Given ab=ca
    if b=c, ba=ca=ab
    The condition for b=c is that G be commutative
    ab=ca is given. If it's true for one a, and b=c, G has to be commutative.

    There are no zero's because every element has an inverse is part of definiton of group. But thanks for noting that anyhow.
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    Re: if ab=ca, how to get b=c???

    Quote Originally Posted by Hartlw View Post
    ab=ca is given. If it's true for one a, and b=c, G has to be commutative.

    There are no zero's because every element has an inverse is part of definiton of group. But thanks for noting that anyhow.
    Let $G = Q_8 = \langle -1,i,j,k \mid (-1)^2 = 1, i^2=j^2=k^2=ijk = -1 \rangle$. Then $Z(G) = \{1,-1\}$. Hence, for any $b,c \in G$, $(-1)b = c(-1)$ implies $b=c$. However, $G$ is not commutative. Example: $ij = k$ but $ji = -k \neq ij$.
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    Re: if ab=ca, how to get b=c???

    So far, it appears that nobody has this right. While it is sufficient that G is abelian, or that a is in Z(G), it is not necessary. All that is required is that a and b commute, or (equivalently) a and c commute, nothing more.

    To see a counter-example, let G = D4, with a = r3, b = c = r. None of these elements are in the center.
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    Re: if ab=ca, how to get b=c???

    Quote Originally Posted by Deveno View Post
    So far, it appears that nobody has this right. While it is sufficient that G is abelian, or that a is in Z(G), it is not necessary. All that is required is that a and b commute, or (equivalently) a and c commute, nothing more.

    To see a counter-example, let G = D4, with a = r3, b = c = r. None of these elements are in the center.
    Good point. I misinterpreted the question. Thank you Deveno.
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    Re: if ab=ca, how to get b=c???

    Quote Originally Posted by Hartlw View Post
    Given ab=ca
    if b=c, ba=ca=ab
    The condition for b=c is that G be commutative
    Specifically, what step do you disagree with?

    What is your answer to OP?
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    Re: if ab=ca, how to get b=c???

    Quote Originally Posted by Hartlw View Post
    Specifically, what step do you disagree with?

    What is your answer to OP?
    As Deveno said, the only condition necessary is that $ab = ba$ (and equivalently $ca = ac$). Obviously, if $G$ is commutative, then $ab = ba$ for all $a,b \in G$. However, there are many non-commutative groups where there exists elements that commute. Both Deveno and I gave examples of such groups. So, while commutativity of $G$ is a sufficient condition, it is not necessary. The only necessary condition is that the specific elements $a$ and $b$ commute.
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    Re: if ab=ca, how to get b=c???

    I didn't prove sufficiency, which is obvious but irrelevant to OP. I proved necessity, which answers OP.
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    Re: if ab=ca, how to get b=c???

    Quote Originally Posted by Hartlw View Post
    I didn't prove sufficiency, which is obvious but irrelevant to OP. I proved necessity, which answers OP.
    Two counterexamples have been posted which prove otherwise (one from me, one from Deveno). I don't know what to tell you. Perhaps you and I are using different definitions for necessary conditions. I am not sure what definition you are using.
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    Re: if ab=ca, how to get b=c???

    Quote Originally Posted by Hartlw View Post
    Given ab=ca
    if b=c, ba=ca=ab
    The condition for b=c is that G be commutative
    For all a.

    The examples were for a specific a.
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