Give an example of group elements $a$ and $b$ with the property that $a^{-1}ba\ne b$. First, I was wondering if it's possible to multiply by $a$ on both left-sides of the equation to obtain $ba \ne ab$? Or is that not allowed because it's ($\ne$) as opposed to $(=)$? Sorry if this is a very trivial question. Secondly, I would like to know if there's a way to figure this out quickly? It seems like I have to apply brute force to find the solution.
 A: Yes, $a^{-1}ba=b \Leftrightarrow ba=ab$ is a consequence of that there is an inverse.
Suppose that you had $a^{-1}ba\ne b$ yet $ba=ab$ then just by multiplying from the left with $a^{-1}$ you reach $a^{-1}ba = b$ which contradicts the assumption. 
Also suppose that $ba\ne ab$ yet $a^{-1}ba = b$ then you use the same technique, but multiply with $a$ from the left and reach $ba=ab$ which contradicts the assumption. 
The only thing that remains is to find an example where $ab\ne ba$, but you probably already know that. Such examples can be found in symmetric group $S_n$ where $n>2$ (take for example in $S_3$ the elements $a=(12)$ and $b=(23)$).
A: 
First, I was wondering if it's possible to multiply by $a$ on both left-sides of the equation to obtain $ba \ne ab$? Or is that not allowed because it's ($\ne$) as opposed to $(=)$?

Yes, you may do this. Specifically it is true in any group that if $x \ne y$, then $ax \ne ay$ and $xa \ne ya$. (This is true because, if $ax$ were equal to $ay$, then we could multiply by $a^{-1}$ to get $x = y$, contradiction.)

Sorry if this is a very trivial question.

"Trivial" is a word that mathematicians use to brag about how smart they are. It means very little.

Secondly, I would like to know if there's a way to figure this out quickly? It seems like I have to apply brute force to find the solution.

Sometimes you just have to have up your sleeve a few of the most simple groups. I mean simply, you should know all the groups of order $2, 3, 4, 6,$ and maybe $8$ and $12$. In this case what you want is $ab \ne ba$, so you are looking for a "nonabelian" group, and there are examples with $6$ elements.
A: One possible group would be a set of all bijections from $[0,1]$ onto $[0,1]$ and group operation to be composition of functions.
Composition of bijections is bijection so we do not leave the structure with this operation.
Inverse of every function exists because of the way bijections are defined.
Neutral element is $e(x)=x$.
Associativity follows from the definition of composition.
Example of non-commutativity can be $f(x)=\sin ( \dfrac {\pi x}{2})$ and $g(x)=\sqrt{x}$.
