Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.14
Prove that $S_4$ does not have a normal subgroup of order 8 or a normal subgroup of order 3.
Solution: Let $K \leq S_4$ be a subgroup of order 3; then $K = \langle (a\ b\ c) \rangle$ for some $(a\ b\ c)$. Note that $$K(a\ d) = \{ (a\ d), (a\ d\ b\ c), (a\ d\ c\ b) \}$$ and $(a\ d)K = \{ (a\ d), (a\ b\ c\ d), (a\ c\ b\ d) \}$, so that $K$ is not normal in $S_4$.
Now suppose $H \leq S_4$ is a subgroup of order 8. Now the order of every element in $H$ must divide 8 by Lagrange, and no element of $S_4$ has order 8. Moreover, $S_4$ has 9 elements of order 2: 6 2-cycles and 3 products of disjoint 2-cycles. Suppose $H$ does not contain an element of order 4. If $H$ contains all 3 products of 2 cycles, then (WLOG) we have $$(a\ c) \circ (a\ b)(c\ d) = (a\ b\ c\ d) \in H,$$ a contradiction. If $H$ does not contain all the products of 2 cycles then it must contain at least 5 2-cycles; WLOG then $H$ contains elements $(a\ b)$, $(a\ c)$, and $(a\ d)$, and $$(a\ d)(a\ c)(a\ b) = (a\ b\ c\ d) \in H,$$ a contradiction. Thus $H$ contains a 4-cycle.
Say $(a\ b\ c\ d) \in H$. Then $\langle (a\ b\ c\ d) \rangle \leq H$, where $$\langle (a\ b\ c\ d) \rangle = \{ 1, (a\ b\ c\ d), (a\ c)(b\ d), (a\ d\ c\ b) \}.$$ Suppose for a moment that $H$ contains another element $\sigma$ of order 4; without loss of generality, $\sigma$ is one of $(a\ b\ d\ c)$ and $(a\ c\ b\ d)$. If $(a\ b\ d\ c) \in H$, then $$\langle (a\ b\ d\ c) \rangle = \{ 1, (a\ b\ d\ c), (a\ d)(b\ c), (a\ c\ d\ b) \} \leq H.$$ But then $(a\ b\ d\ c)(a\ b\ c\ d) = (a\ d\ b) \in H$, but $(a\ d\ b)$ has order 3. Thus $(a\ b\ d\ c) \notin H$. Similarly, $(a\ c\ b\ d) \notin H$, hence the remaining elements of $H$ all have order 2. There are 8 such elements remaining.
Note that $$(a\ b)(a\ b\ c\ d) = (b\ c\ d),$$ $$(a\ d)(a\ b\ c\ d) = (a\ b\ c),$$ $$(b\ c)(a\ b\ c\ d) = (a\ c\ d),$$ and $$(c\ d)(a\ b\ c\ d) = (a\ b\ d),$$ so that $(a\ b)$, $(a\ d)$, $(b\ c)$, and $(c\ d)$ are not in $H$. Then the remaining elements must be in $H$, so that $$H = \{ 1, (a\ b\ c\ d), (a\ c)(b\ d), (a\ d\ c\ b), (a\ c), (b\ d), (a\ b)(c\ d), (a\ d)(b\ c) \}.$$ A simple calculation shows that indeed $H = \langle (a\ b\ c\ d), (b\ d) \rangle$.
We saw in the previous exercise that, if $K = \langle (a\ b\ c) \rangle$, then $HK \neq KH$. By Corollary 15 in the text, then, $H$ is not normal in $S_4$.
