<s>[INST] 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 $R = PQ$?

 [/INST] 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
$$PQ^{-1}Q=P.$$
</s>