**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.5 Exercise 4.5.7**

Solution: The Sylow 2-subgroups of $S_4$ have order 8, and by Sylow’s Theorem, $n_2 = 1 \pmod 2$ and $n_2$ divides 3. Thus $n_2 \in \{1,3\}$.

In Exercise 4.2.5, we found that for any bijection $\{a,b,c,d\} \rightarrow \{1,2,3,4\}$, $\langle (a\ b\ c\ d), (b\ d) \rangle$ is a subgroup of order 8 in $S_4$.

Note that $$H_1 = \langle (1\ 2\ 3\ 4), (2\ 4) \rangle = \{ 1, (1\ 2\ 3\ 4), (1\ 3)(2\ 4), (1\ 4\ 3\ 2), (2\ 4), (1\ 4)(2\ 3), (1\ 3), (1\ 2)(3\ 4) \}$$ $$H_2 = \langle (1\ 3\ 2\ 4), (3\ 4) \rangle = \{ 1, (1\ 3\ 2\ 4), (1\ 2)(3\ 4), (1\ 4\ 2\ 3), (3\ 4), (1\ 4)(2\ 3), (1\ 2), (1\ 3)(2\ 4) \}$$ $$H_3 = \langle (1\ 2\ 4\ 3), (2\ 3) \rangle = \{ 1, (1\ 2\ 4\ 3), (1\ 4)(2\ 3), (1\ 3\ 4\ 2), (2\ 3), (1\ 3)(2\ 4), (1\ 4), (1\ 2)(3\ 4) \}$$ are distinct subgroups of order 8 in $S_4$, so that we have exhausted the full list of Sylow 2-subgroups of $S_4$.

Note that $$(2\ 3) H_1 (2\ 3) = H_2, (3\ 4) H_1 (3\ 4) = H_3,$$ and $(3\ 2\ 4) H_2 (3\ 4\ 2) = H_3$.