Question

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

Answer 1

Answer:

The answer is $x=a^{-1}cb^{-1}$.

Step-by-step explanation:

First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element $g\in G$ there exist another element, denoted by $g^{-1}$ such that $gg^{-1}=e$, where $e$ stands for the identity element of G.

So, given the equality $axb=c$ we make a left multiplication by $a^{-1}$ and we obtain:

$a^{-1}axb =a^{-1}c.$

But, $a^{-1}axb = exb = xb$. Hence, $xb = a^{-1}c$.

Now, in the equality $xb = a^{-1}c$ we make a right multiplication by $b^{-1}$, and we obtain

$xbb^{-1} = a^{-1}cb^{-1}$.

Recall that $bb^{-1}=e$ and $xe=x$. Therefore,

$x=a^{-1}cb^{-1}$.