Question

In a group G, the equations ax = b and xa = b have,
A. no solution in G.
B. infinite solution in G.
C. unique solution in G.
D. depends on a and b.

Answer 1

Hint: Prove this using the elementary properties of groups. The given question is part of a theorem. Assume basic properties of any group G and find the type of solution formed.

Complete step by step solution:
This question deals with the elementary properties of a group.
For any group G, let us assume the following properties.
If a, b, c\[\in \]G and ab = ac then b = c, by left cancellation law.
If a, b, c\[\in \]G and ba = ca then b =c, by right cancellation law.
If a\[\in \]G then, \[{{\left( {{a}^{-1}} \right)}^{-1}}=a\]. The inverse of the inverse of an element is the element itself.
If a\[\in \]G, then, \[a{{a}^{-1}}={{a}^{-1}}.a=e\], then a is an inverse of \[{{a}^{-1}}\].
Since, inverses are unique then \[{{\left( {{a}^{-1}} \right)}^{-1}}=a\].
Now, we know the basic properties.
In the given group G, a, b \[\in \]G. Hence, we need to find the solution of equations ax = b and xa = b.
Let us first take, \[ax=b\].
\[\begin{align}
  & \therefore x=\dfrac{b}{a}={{a}^{-1}}b \\
 & x={{a}^{-1}}b \\
\end{align}\]
Hence, we have isolated x on the left side or it can be done by taking\[\Rightarrow x=ex\].
We got \[e={{a}^{-1}}a\]from the properties.
\[\therefore x=\left( {{a}^{-1}}a \right)x\] (we have ax = b)
\[\begin{align}
  & x={{a}^{-1}}\left( ax \right) \\
 & x={{a}^{-1}}b \\
\end{align}\]
Now let us take, xa= b.
\[\begin{align}
  & \therefore x=\dfrac{b}{a}={{a}^{-1}}b \\
 & x={{a}^{-1}}b \\
\end{align}\]
Thus by using the same properties above we can say that \[\Rightarrow x=ex\].
We know, \[e={{a}^{-1}}a\].
\[\therefore x=\left( {{a}^{-1}}a \right)x\] (we have xa = b)
Thus we got,
 \[\begin{align}
  & a={{a}^{-1}}\left( ax \right) \\
 & x={{a}^{-1}}b \\
\end{align}\]
Now, let us assume that y is another solution to the equation ax = b. Then, ay = b =ax.
By the left cancellation property, we have x = y.
Hence, proof is that ax = b is similar to xa = b.
Hence, in a group G, a, b\[\in \]G then each of equations ax = b and xa = b has a unique solution in G.
Hence, option (c) is the correct answer.

Note: The equation ax = b has a unique solution means that b appears only once in the row of a Cayley table which describes the structure of a finite group by arranging all possible products of the group. Similarly, there is a unique solution for xa = b, and each element appears at exactly once in each column of the table. Hence ax =b and xa = b have a unique solution.