3

From Prof. Charles Pinter's A Book of Abstract Algebra's Chapter 4 exercises:

Let $a, b, c$ and $x$ be elements of a group $G$. In each of the following, solve for $x$ in terms of $a, b$ and $c$.

Problem 1: $$axb = c$$

I came up with an answer:

$$x = c/ab$$

i.e.

$$x = ca^{-1}b^{-1}$$

But I'm not sure if this answer is correct. It seems somewhat simple given that there's one operation (divide each side by $ab$) to get the answer.

Please comment on my answer. If I'm right, please explain why there's no further break-down/simplification. If I'm wrong, please guide me.

Community
  • 1
Kevin Meredith
  • 969
  • What does $c/ab$ mean? There is no $/$ operation in a group. – anon May 13 '15 at 02:45
  • 3
    remember that multiplication is not necessarily commutative – yoyo May 13 '15 at 02:47
  • If you invert both sides you get $(axb)^{-1}=c^{-1}$. You may not be familiar with how just yet, but it turns out that $(axb)^{-1}$ can be simplified to $b^{-1}x^{-1}a^{-1}$. So your equation becomes $b^{-1}x^{-1}a^{-1}=c^{-1}$, and you are no closer to solving for $x$. – anon May 13 '15 at 02:49

3 Answers3

2

There is no $/$ operation in a group; the expression $c/ab$ literally does not mean anything.

What operations do you have available to you for manipulating equations involving elements from the group? You can invert both sides of an equation, you can multiply both sides on the left by an element of your choosing, and you can multiply both sides on the right by an element of your choosing.

Given $axb=c$, how do you get rid of the $a$ in front of $x$, and how do you get rid of $b$ behind the $x$?

anon
  • 151,657
2

To check whether your answer is correct or not, just plug into the equation the value of $x$ you've found and see if equality holds:

$$axb=a(ca^{-1}b^{-1})b=(aca^{-1})(b^{-1}b)=aca^{-1}$$

but this last value might not be $c$ (if the group is abelian then yes, it equals $c$, but that doesn't hold necesarily if the group is not abelian).

Daniel
  • 6,999
1

Well, if they don't specify that the group $G$ is abelian, you can't just "divide" that way, since the operation may be non-commutative. What you would have to do is right- or left-multiply both sides of the expression by $a^{-1}$ and $b^{-1}$. So here, you would start by right-multiplying by $b^{-1}$, then you would have $ax=cb^{-1}$, which you could then left-multiply by $a^{-1}$ and get $x=a^{-1}cb^{-1}$.

JoDraX
  • 258
  • 1
    http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Daniel May 13 '15 at 02:52