Have seen a youtube video on composition of elements of $D_4$, symmetries of a square.
At 8.33 and before, a composition $\delta_2(\rho_3(1234))$ is given. The output differs than that obtained by me below.
Request to tell why difference occurs.
$$\delta_2= (1423), \rho_3=(1432)$$ $$\delta_2= \begin{pmatrix} 1 & 2 & 3 &4 \\ 4 & 1 & 3 & 2\end{pmatrix} \implies (1423)$$ $$\rho_3= \begin{pmatrix} 1 & 2 & 3 &4 \\ 4 & 1 & 2 & 3\end{pmatrix} \implies (1432)$$ First mapping by $\rho_3: (1432)$. Next, by $\delta_2=(1423)$ as below: $1_1\to 4_1, 4_2 \to 2_2$ $2_1\to 1_1, 1_2 \to 4_2$ $3_1\to 2_1, 2_2 \to 1_2$ $4_1\to 3_1, 3_2 \to 3_2$
Or, $(1243)$
But, seems from the video at 8:33, that the output of composition is two disjoint cycles with $1\to 2, 2 \to 1, 3\to4, 4\to 3$, or $$= \begin{pmatrix} 1 & 2 & 3 &4 \\ 2 & 1 & 4 & 3\end{pmatrix}\implies (12)(34)$$
Edit :
There is mistake in having one fixed point in $\delta_2$:
$$\delta_2= (1)(42)(3), \rho_3=(1432)$$ $$\delta_2= \begin{pmatrix} 1 & 2 & 3 &4 \\ 1 & 4 & 3 & 2 \end{pmatrix}$$ $$\rho_3= \begin{pmatrix} 1 & 2 & 3 &4 \\ 4 & 1 & 2 & 3\end{pmatrix} \implies (1432)$$ First mapping by $\rho_3: (1432)$. Next, by $\delta_2=(42)$ as below: $1_1\to 4_1, 4_2 \to 2_2$ $2_1\to 1_1, 1_2 \to 1_2$ $3_1\to 2_1, 2_2 \to 4_2$ $4_1\to 3_1, 3_2 \to 3_2$
The composition yields correct result now $$= \begin{pmatrix} 1_1 & 2_1 & 3_1 &4_1 \\ 2_2 & 1_2 & 4_2 & 3_2 \end{pmatrix}\implies (12)(34)$$
Next have correctly given the next (reverse) composition in the video, as below:
$$\delta_2= (1)(42)(3), \rho_3=(1432)$$ $$\delta_2= \begin{pmatrix} 1 & 2 & 3 &4 \\ 1 & 4 & 3 & 2 \end{pmatrix}$$ $$\rho_3= \begin{pmatrix} 1 & 2 & 3 &4 \\ 4 & 1 & 2 & 3\end{pmatrix} $$ $1_1\to 1_1, 1_2 \to 4_2$ $2_1\to 4_1, 4_2 \to 3_2$ $3_1\to 3_1, 3_2 \to 2_2$ $4_1\to 2_1, 2_2 \to 1_2$ $$\rho_3(\delta_2(1234))= \begin{pmatrix} 1 & 2 & 3 &4 \\ 4 & 3 & 2 & 1\end{pmatrix} \implies (14)(32)$$