Let $\displaystyle a, b, c \in G$ where $\displaystyle G$ is a finite group.

If $\displaystyle ab=ca$, what conditions we need in order to have $\displaystyle b=c$???

conjugacy class?? order??

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- Feb 17th 2014, 06:27 AMdeniselim17if ab=ca, how to get b=c???
Let $\displaystyle a, b, c \in G$ where $\displaystyle G$ is a finite group.

If $\displaystyle ab=ca$, what conditions we need in order to have $\displaystyle b=c$???

conjugacy class?? order?? - Feb 17th 2014, 05:06 PMchiroRe: 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). - Feb 18th 2014, 06:29 AMHartlwRe: 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 - Feb 18th 2014, 10:02 AMebainesRe: if ab=ca, how to get b=c???
- Feb 18th 2014, 01:46 PMSlipEternalRe: 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 $\displaystyle a \neq 0$. 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 $\displaystyle a \in Z(G)$ where $\displaystyle Z(G) = \{z \in G \mid \forall g\in G, zg = gz\}$ is the center of the group.

- Feb 18th 2014, 01:55 PMromsekRe: if ab=ca, how to get b=c???
- Feb 19th 2014, 05:41 AMHartlwRe: if ab=ca, how to get b=c???
- Feb 19th 2014, 06:07 AMSlipEternalRe: if ab=ca, how to get b=c???
- Feb 19th 2014, 06:17 AMDevenoRe: 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. - Feb 19th 2014, 06:27 AMSlipEternalRe: if ab=ca, how to get b=c???
- Feb 19th 2014, 07:10 AMHartlwRe: if ab=ca, how to get b=c???
- Feb 19th 2014, 07:18 AMSlipEternalRe: 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. - Feb 19th 2014, 07:41 AMHartlwRe: 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.

- Feb 19th 2014, 07:54 AMSlipEternalRe: if ab=ca, how to get b=c???
- Feb 19th 2014, 08:07 AMHartlwRe: if ab=ca, how to get b=c???