Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.5
Solution: To save effort, we compute the representation of $D_8$ in $S_{D_8}$.$$1 \mapsto 1$$ $$r(H) = rH, r(rH) = r^2H, r(r^2H) = r^3H, r(r^3H) = H.$$ Thus $r \mapsto (H\ rH\ r^2H\ r^3H)$. $$r^2(H) = r^2H r^2(rH) = r^3H, r^2(r^2H) = H, r^2(r^3H) = rH.$$ Thus $r^2 \mapsto (H\ r^2H)(rH\ r^3H)$. $$r^3(H) = r^3H, r^3(rH) = H, r^3(r^2H) = rH, r^3(r^3H) = r^2H.$$ Thus $r^3 \mapsto (H\ r^3H\ r^2H\ rH)$. $$s(H) = sH = H, s(rH) = srH = r^3sH = r^3H,$$ $$s(r^2H) = sr^2H = r^2sH = r^2H,$$ and $$s(r^3H) = sr^3H = rsH = rH.$$ Thus $s \mapsto (rH\ r^3H)$. $$sr(H) = srH = r^3sH = r^3H,$$ $$sr(rH) = sr^2H = r^2sH = r^2H,$$ $$sr(r^2H) = sr^3H = rsH = rH,$$ and $$sr(r^3H) = sH = H.$$ Thus $sr \mapsto (H\ r^3H)(rH\ r^2H)$. $$sr^2(H) = sr^2H = r^2sH = r^2H,$$ $$sr^2(rH) = sr^3H = rsH = rH,$$ $$sr^2(r^2H) = sH = H,$$ and $$sr^2(r^3H) = srH = r^3sH = r^3H.$$ Thus $sr^2 \mapsto (H\ r^2H)$. $$sr^3(H) = sr^3H = rsH = rH,$$ $$sr^3(rH) = sH = H, sr^3(r^2H) = srH = r^3sH = r^3H,$$ and $$sr^3(r^3H) = sr^2H = r^2sH = r^2H.$$ Thus $sr^3 \mapsto (H\ rH)(r^2H\ r^3H)$.
Clearly this is a faithful representation of $D_8$ in $S_4$.
Now under the bijection$$H \mapsto 1\quad rH \mapsto 2\quad r^2H \mapsto 3\quad r^3H \mapsto 4$$ we have $$D_8 \cong \langle (1\ 2\ 3\ 4), (2\ 4) \rangle. $$Under the bijection$$H \mapsto 1\quad rH \mapsto 3\quad r^2H \mapsto 2 \quad r^3H \mapsto 4$$ we have $$D_8 \cong \langle (1\ 3\ 2\ 4), (3\ 4) \rangle.$$ Now if these subgroups are equal, then we have $(2\ 4)(3\ 4) = (2\ 4\ 3)$ contained in an isomorphic copy of $D_8$. But $(2\ 4\ 3)$ has order 3, a contradiction since $D_8$ has order 8. Thus these two subgroups are distinct.
Recall that $D_8 = \langle r, sr^3 \rangle$; thus the image of $D_8$ in $S_4$ via the action on $K$ by left multiplication is generated by the images of $r$ and $sr^3$.
Now $$r(K) = rK, r(rK) = r^2K, r(r^2K) = r^3K, r(r^3K) = K,$$ so that $r \mapsto (K\ rK\ r^2K\ r^3K)$. Similarly, $$sr^3(K) = sr^4sK = K, sr^3(rK) = sK = srsK = r^3K,$$ $$sr^3(r^2K) = srK = K, sr^3(r^3K) = sr^2K = rrsK = rK.$$ Thus $sr^3 \mapsto (rK\ r^3K)$. With the labeling $$K \mapsto 1, rK \mapsto 2, r^2K \mapsto 3, r^3K \mapsto 4,$$ we have $$\mathsf{im}\ D_8 = \langle (1\ 2\ 3\ 4), (2\ 4) \rangle$$ as desired.
