$SS^x = S^x S$ implies that $S$ is subnormal Suppose $S$ is a subgroup of a finite group $G$, and suppose that $SS^x = S^x S$ for all $x \in G$. In M.Isaacs - finite group theory, the way of proving that this implies $S$ is subnormal in $G$ is to first prove the zipper lemma and then use it.
I thought of an alternative approach that doesn't use the zipper lemma, and I'd be glad if someone checked whether it makes sense. It's very similar to the approach in the book - use induction on order of $G$, and consider the closure $S^G$. But instead of using the zipper lemma, I attempt to directly show that if $G=S^G$, then $S=G$, using $SS^x = S^x S$ and induction.
Notice that from $SS^x = S^x S$ for all $x \in G$ it follows that $S^a S^b = S^b S^a$ for all $a,b \in G$. Choose some set of elements $\{ x_1,..,x_n \}$ such that $\langle S^{x_1},...,S^{x_n} \rangle = S^G = \langle S^x : x \in G \rangle$ (we are using finiteness of $G$ here).
We will now check that $S^{x_1}\cdot \cdot \cdot S^{x_n} = S^G$. The $\subseteq$ inclusion is clear. To prove $\supseteq$, suppose that $\alpha \in S^G$. Then $\alpha$ can be expressed as $\alpha_{i_1} \cdot \cdot \cdot \alpha_{i_m}$ where each $\alpha_{i_j}$ is contained in $S^{i_j} \in  \{ S^{x_1},...,S^{x_n} \}$, and so $\alpha \in S^{i_1} \cdot \cdot \cdot S^{i_m}$. Using the equality $S^a S^b = S^a S^b$ described in the previous paragraph, we see that $S^{i_1} \cdot \cdot \cdot S^{i_m} \subseteq S^{x_1} \cdot \cdot \cdot S^{x_n}$.
We now proceed to showing $S$ is subnormal in $G$ by induction on the order of $G$. By the induction hypothesis, if $S \leq H \neq G$, then $S$ is subnormal in $H$. If $S^G \neq G$, then $S$ is subnormal in $S^G$ and $S^G$ is normal in $G$, so we are done. 
If $S^G = G$, then by the previous paragraph we know that $S^{x_1}\cdot \cdot \cdot S^{x_n} = G$. Now it is generally true that if $AB = G$ for subgroups $A,B$, then $A^u B^v = G$ for any $u,v \in G$ - to see this, find $ab \in AB$ so that $uv^{-1} = ab$, then
\begin{align*}
u^{-1} A uv^{-1} B v = u^{-1} AabB v = u^{-1} AB v = u^{-1}Gv =u^{-1} G = G.
\end{align*}
Using this observation on $S^{x_1}\cdot \cdot \cdot S^{x_n} = G$, we can describe $G$ as 
\begin{align*}
(S^{x_1} ) ( S^{x_2} \cdot \cdot \cdot S^{x_n}) = (S^{x_1})^{{x_1}^{-1} x_2} ( S^{x_2} \cdot \cdot \cdot S^{x_n}) = S^{x_2} ( S^{x_2} \cdot \cdot \cdot S^{x_n}) = ( S^{x_2} \cdot \cdot \cdot S^{x_n}),
\end{align*}
 so by induction, we see that $S^{x_n}=G$, which implies $S = G$, so $S$ is subnormal in G$ as well.
 A: The problem is in the last few lines: you want to apply the rule that $AB = G$ implies $A^uB^v = G$ to $A = S^{x_1}$ and $B = S^{x_2}\cdots S^{x_n}$, but there is no reason in general that $S^{x_2}\cdots S^{x_n}$ is a subgroup of $G$. 
Another remark is that the existence of $x_1, \ldots, x_n \in G$ with $S^{x_1}\cdots S^{x_n} = G$ already follows from $S^G = G$ and finiteness of $G$, without any hypothesis on $S$. To see this, write $G = \{g_1, \ldots, g_m\}$. For each $j$, we have $g_j \in S^G$, and so $g_j$ can be written as a product of elements of conjugates of $S$. In other words, there exist $x_{j,1}, \ldots, x_{j,k_m} \in G$ (not necessarily distinct) such that $g_j \in S^{x_{j,1}}\cdots S^{x_{j,k_m}}$. Taking products of subsets that contain the identity only makes them bigger, so we find that 
$$ G = S^{x_{1,1}} \cdots S^{x_{1, k_1}}S^{x_{2, 1}}\cdots S^{x_{2,k_2}} \cdots\cdots S^{x_{m,1}} \cdots S^{x_{m, k_m}} $$
since this product by construction contains all elements of $G$. (More generally, this argument shows that if $S^G$ is finite, then there is a finite list $x_1, \ldots, x_n \in G$ with $S^G = S^{x_1}\cdots S^{x_n}$.)
So you see that, effectively, the last paragraph never uses the hypothesis that $S^x S = SS^x$ for $x \in G$. Unfortunately, without this hypothesis the statement that you want to prove, namely that $S^G = G$ for $G$ finite implies $S$ subnormal in $G$, is simple not true. For a general sort of counterexample, take $G$ non-abelian finite simple and $S$ any non-trivial proper subgroup of $G$. 
