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

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• February 17th 2014, 07:27 AM
deniselim17
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??
• February 17th 2014, 06:06 PM
chiro
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
Hartlw
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
• February 18th 2014, 11:02 AM
ebaines
Re: if ab=ca, how to get b=c???
Quote:

Originally Posted by Hartlw
The condition for b=c is that G be commutative

And that $a \ne 0$.
• February 18th 2014, 02:46 PM
SlipEternal
Re: if ab=ca, how to get b=c???
Quote:

Originally Posted by ebaines
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.
• February 18th 2014, 02:55 PM
romsek
Re: if ab=ca, how to get b=c???
Quote:

Originally Posted by SlipEternal
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
• February 19th 2014, 06:41 AM
Hartlw
Re: if ab=ca, how to get b=c???
Quote:

Originally Posted by Hartlw
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.
• February 19th 2014, 07:07 AM
SlipEternal
Re: if ab=ca, how to get b=c???
Quote:

Originally Posted by Hartlw
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 $ji = -k \neq ij$.
• February 19th 2014, 07:17 AM
Deveno
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.
• February 19th 2014, 07:27 AM
SlipEternal
Re: if ab=ca, how to get b=c???
Quote:

Originally Posted by Deveno
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.
• February 19th 2014, 08:10 AM
Hartlw
Re: if ab=ca, how to get b=c???
Quote:

Originally Posted by Hartlw
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?

• February 19th 2014, 08:18 AM
SlipEternal
Re: if ab=ca, how to get b=c???
Quote:

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

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
Hartlw
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.
• February 19th 2014, 08:54 AM
SlipEternal
Re: if ab=ca, how to get b=c???
Quote:

Originally Posted by Hartlw
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.
• February 19th 2014, 09:07 AM
Hartlw
Re: if ab=ca, how to get b=c???
Quote:

Originally Posted by Hartlw
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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