Solve the equation $a \circ x \circ b = c$ in the group $S_4$

Question

Solve the equation $a \circ x \circ b = c$ in the group $S_4$ when $$a=\begin{pmatrix}1&2&3&4\\ 2&3&1&4\end{pmatrix}, b= \begin{pmatrix}1&2&3&4\\ 4&1&2&3\end{pmatrix}, c=\begin{pmatrix}1&2&3&4\\ 1&3&4&2\end{pmatrix}.$$

So going from right to left in $a \circ x \circ b$ we want that $$\begin{align*} 1 \to 4 \to 3 \to 1\end{align*} \\ 2 \to 1 \to 2 \to 3 \\ 3 \to 2 \to 4 \to 4 \\ 4 \to 3 \to 1 \to 2$$ this gave me that $x =\begin{pmatrix}1&2&3&4\\ 2&4&1&3\end{pmatrix}$ but I'm not sure how I can verify this?

Using cycle notation this seems to be false as $$(123)\circ x \circ (1432)$$ would be then $$(123)(1243)(1432)=(13)(24)$$ and not $(234)$.

Answer 1

Your answer is correct and, actually, we don't have $$(1\ \ 2\ \ 3)\circ(1\ \ 2\ \ 4\ \ 3)\circ(1\ \ 4\ \ 3\ \ 2)=(1\ \ 3)\circ(2\ \ 4).\tag1\label{1}$$ For instance, the LHS of \eqref{1} maps $1$ into $1$, since:

• $(1\ \ 4\ \ 3\ \ 2)$ maps $1$ into $4$;
• $(1\ \ 2\ \ 4\ \ 3)$ maps $4$ into $3$;
• $(1\ \ 2\ \ 3)$ maps $3$ into $1$.

Therefore, \eqref{1} is false. And, in fact, $$(1\ \ 2\ \ 3)\circ(1\ \ 2\ \ 4\ \ 3)\circ(1\ \ 4\ \ 3\ \ 2)=(2\ \ 3\ \ 4)=c.$$

Answer 2

You can solve it algebraically using the inverse transformations:

$x=a^{-1}\circ c\circ b^{-1}=\begin{pmatrix}1&2&3&4\\3&1&2&4\end{pmatrix}\circ\begin{pmatrix}1&2&3&4\\1&3&4&2\end{pmatrix}\circ\begin{pmatrix}1&2&3&4\\2&3&4&1\end{pmatrix}=\begin{pmatrix}1&2&3&4\\2&4&1&3\end{pmatrix}$