**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.6**

Solution: We compute the permutation representation of $G$ in $D_{G/N}$.$$1 \mapsto 1$$

$$r(N) = rN, r(rN) = N, r(sN) = rsN = sr^3N = srN, r(srN) = sr^4N = sN.$$ Thus $r \mapsto (N\ rN)(sN\ srN)$. $$r^2(N) = N, r^2(rN) = rN, r^2(sN) = sr^2N = sN, r^2(srN) = sr^3N = srN.$$ Thus $r^2 \mapsto 1$. $$r^3(N) = rN, r^3(rN) = N, r^3(sN) = srN, r^3(srN) = sN.$$ Thus $r^3 \mapsto (N\ rN)(sN\ srN)$. $$s(N) = sN, s(rN) = srN, s(sN) = N, s(srN) = rN.$$ Thus $s \mapsto (N\ sN)(rN\ srN)$. $$sr(N) = srN, sr(rN) = sN, sr(sN) = r^3N = rN, sr(srN) = N.$$ Thus $sr \mapsto (N\ srN)(rN\ sN)$. $$sr^2(N) = sN, sr^2(rN) = srN, sr^2(sN) = N, sr^2(srN) = rN.$$ Thus $sr^2 \mapsto (N\ sN)(rN\ srN)$. $$sr^3(N) = srN, sr^3(rN) = sN, sr^3(sN) = rN, sr^3(srN) = N.$$ Thus $sr^3 \mapsto (N\ srN)(rN\ sN)$. So the image of this representation is $$\{1, (N\ rN)(sN\ srN), (N\ sN)(rN\ srN), (N\ srN)(rN\ sN) \}.$$ With the bijection $N \mapsto 1$, $rN \mapsto 2$, $sN \mapsto 3$, and $srN \mapsto 4$, this is the group $\langle (1\ 2)(3\ 4), (1\ 3)(2\ 4) \rangle$.

This is a group of order 4 with no element of order 4; thus the image of $D_8$ is isomorphic to $V_4$.