Question: 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 $R = PQ$?

[![enter image description here](https://i.stack.imgur.com/HRFfS.jpg)](https://i.stack.imgur.com/HRFfS.jpg)

Answer: 
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

$$PQ^{-1}Q=P.$$
> 
> 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 $\operatorname {cl}(P):=\{Q^{-1}PQ, Q\in G\}$.