Group theory, when are elements conjugate to each other? In my book "Group Theory and Chemistry" by Bishop, it is written that if $$R = Q^{-1}PQ$$ $R$ is the transform of $P$ by $Q.$ But isn't $R$ not just equal to $P (Q^{-1} Q = E )$?
If I transform $P$ by $Q$ wouldn't I have $R = PQ$?

 A: I will give two examples where this fails. The first is the smallest possible group that is non-abelian; that is, where the elements do not necessarily commute, so there exist $a,b$ in the group such that $ab\neq ba$. The second is usually one of the first examples one studies where commutativity fails.

Consider the group $D_3$ of all rotational and reflectional symmetries of an equilateral triangle.
Label the vertices of the triangle (pointing up) $1$, $2$, and $3$, going clockwise from the top.
Let $P$ be the clockwise rotation, sending $1$ to $2$, $2$ to $3$, and $3$ to $1$. Then $P^{-1}$ is the counter-clockwise rotation, sending, in the original configuration,  $1$ to $3$, $3$ to $2$, and $2$ to $1$.
Let $Q$ be the vertical reflection, fixing $1$ and sending $2$ to $3$ and $3$ to $2$. Then clearly $Q^{-1}$ is $Q$.
Define concatenation $AB$ of symmetries $A,B$ as doing $B$ then $A$.
We have
$$Q^{-1}PQ=P^{-1},$$
whereas
$$PQ^{-1}Q=P.$$

Consider $GL_2(\Bbb R)$, the group of invertible $2\times 2$ matrices with entries in $\Bbb R$.
Consider
$$Q=\begin{pmatrix}
0 & 1\\
1 & -1
\end{pmatrix}.$$
Then
$$Q^{-1}=\begin{pmatrix}
1 & 1\\
1 & 0
\end{pmatrix}.$$
Let
$$P=\begin{pmatrix}
1 & 0\\
1 & 0
\end{pmatrix}.$$
Then
$$\begin{align}
Q^{-1}PQ&=\begin{pmatrix}
1 & 1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1\\
1 & -1
\end{pmatrix}
\\
&=\begin{pmatrix}
2 & 0\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1\\
1 & -1
\end{pmatrix}
\\
&=\begin{pmatrix}
0 & 2\\
0 & 1
\end{pmatrix},
\end{align}$$
whereas
$$PQ^{-1}Q=P.$$
A: The group operation is not necessarily commutative. $Q^{-1} P Q$ is different from $P Q^{-1} Q.~$ A conjugate is the most common kind of transformation in a group, so sometimes it is called the transform.
A: I like to think of a concrete example. Let $P$ be the rotation of the plane by $90^\circ$ centred at some point $A$, and let $Q$ be the translation of the plane which maps $A$ to another point $B$. Then reading $Q^{-1}PQ$ left-to-right (ie think of these transformations acting on the right), this composite transform maps $B$ to $A$, then rotates about $A$, and finally translates $A$ back to $B$. Overall then, $R=Q^{-1}PQ$ is the rotation of the plane by $90^\circ$ about the point $B$.
$P$ and $R$ are the same 'type' of transformation, ie rotations by $90^\circ$, but $Q$ transforms $P$ into $R$ in that the centre of rotation shifts from $A$ to $B$.
A: 
If I transform $P$ by $Q$ wouldn't I have $R = PQ$?

Well, this is just another way of transforming $P$ by $Q$, and it's quite informative for every group (Abelian and not). In fact, the map $Q\mapsto PQ$ is a bijection from the group to itself (namely a member of group's symmetric group), and the map $P\mapsto(Q\mapsto PQ)$ is an injective group homomorphism from the group to its symmetric group. So, for every group, its symmetric group contains an isomorphic copy of the group itself (Cayley's theorem).
On the other hand, for a non-Abelian group $G$, the transformation $Q\mapsto Q^{-1}PQ$ is a surjective one from the group to the (eventually) non-trivial conjugacy class of $P$, which is indeed defined by $\operatorname {cl}(P):=\{Q^{-1}PQ, Q\in G\}$.
