Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.4
Solution: Recall that $Q_8 = \langle i,j \rangle$.
Now $i(1) = i$, $i(-1) = -i$, $i(i) = -1$, $i(-i) = 1$, $i(j) = k$, $i(-j) = -k$, $i(k) = -j$, and $i(-k) = j$. Thus $$i \mapsto (1\ i\ -1\ -i)(j\ k\ -j\ -k).$$ Similarly, $j(1) = j$, $j(-1) = -j$, $j(i) = -k$, $j(-i) = k$, $j(j) = -1$, $j(-j) = -1$, $j(k) = i$, and $j(-k) = -i$. Thus $$j \mapsto (1\ j\ -1\ -j)(i\ -k\ -i\ k).$$With the labeling $$1 \mapsto 1 \quad -1 \mapsto 2\quad i \mapsto 3 \quad -i \mapsto 4$$ $$j \mapsto 5 \quad -j \mapsto 6 \quad k \mapsto 7 \quad -k \mapsto 8$$ we have $$Q_8 \cong \langle (1\ 3\ 2\ 4)(5\ 7\ 6\ 8), (1\ 5\ 2\ 6)(3\ 8\ 4\ 7) \rangle.$$
