Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.5 Exercise 4.5.8
Solution: $|S_5| = 120 = 2^3 \cdot 3 \cdot 5$, so that the Sylow 2-subgroups of $S_5$ have order 8. Note that $S_4$ is canonically a subgroup of $S_5$, and that as we saw in Exercise 4.5.7, the Sylow 2-subgroups of $S_4$ have order 8. Thus every Sylow 2-subgroup in $S_4$ is (canonically) a Sylow 2-subgroup in $S_5$.
In Exercise 4.5.7, we found that $$H_1 = \langle (1\ 2\ 3\ 4), (2\ 4) \rangle$$ and $$H_2 = \langle (1\ 3\ 2\ 4), (3\ 4) \rangle$$ are distinct Sylow 2-subgroups of $S_4$, hence distinct Sylow 2-subgroups of $S_5$. Moreover, $(2\ 3) H_1 (2\ 3) = H_2$.
