Let where is a finite group.

If , what conditions we need in order to have ???

conjugacy class?? order??

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- February 17th 2014, 07:27 AM #1

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- February 17th 2014, 06:06 PM #2

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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).

- February 18th 2014, 07:29 AM #3

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- February 18th 2014, 11:02 AM #4

- February 18th 2014, 02:46 PM #5

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## Re: if ab=ca, how to get b=c???

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 . 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 where is the center of the group.

- February 18th 2014, 02:55 PM #6

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- February 19th 2014, 06:41 AM #7

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## Re: if ab=ca, how to get b=c???

- February 19th 2014, 07:07 AM #8

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## Re: if ab=ca, how to get b=c???

- February 19th 2014, 07:17 AM #9

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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 = D_{4}, with a = r^{3}, b = c = r. None of these elements are in the center.

- February 19th 2014, 07:27 AM #10

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- February 19th 2014, 08:10 AM #11

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- February 19th 2014, 08:18 AM #12

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## Re: if ab=ca, how to get b=c???

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.

- February 19th 2014, 08:41 AM #13

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- February 19th 2014, 08:54 AM #14

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## Re: if ab=ca, how to get b=c???

- February 19th 2014, 09:07 AM #15

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