Does intersection of subgroups preserve the property of being generated by transpositions? I'm reading a book and the author seems to just assume this proposition at one point, but upon trying it myself I don't see why it's the case, even though I can't find a counterexample.
Statement:
Let $U = U_1 \cap U_2$, where $U_1, U_2 \leq S_n$, the symmetric group, and $U_1, U_2$ are generated by transpositions. Then $U$ is also generated by transpositions.
The author assumes it in the case where $U_1$ is generated by transpositions of the form $(i, i+1)$, so at least that case should be true.
I don't really know where to start with this problem.
Writing $g = s_1s_2\cdots s_k = t_1t_2 \cdots t_l$, where $s_i \in U_1, t_i \in U_2$ are transpositions, I don't really know how to continue in order to find an expression of $g$ in terms of transpositions in $U_1 \cap U_2$.
A more general version of the statement that I also don't know whether it's true is:
Statement:
Let $U = U_1 \cap U_2$, where $U_1, U_2 \leq G$, some larger group, and let $T$ be a subset of $G$. Suppose $U_1$ is generated by $U_1\cap T$ and $U_2$ is generated by $U_2\cap T$. Then $U$ is generated by $U_1\cap U_2 \cap T$.
I don't see how to start with this one either.
 A: (Here I prove that the intersection of two subgroups of a symmetric group that are generated by transpositions is itself generated by transpositions.)
Let $G$ be a subgroup of $S_n$ generated by transpositions. We claim that $G$ is a product of the symmetric groups on the orbits of $G$. It suffices to do this in the case where $G$ is transitive, and here we may assume that $(1,2)$ lies in $G$. Let $\sim$ be the relation given by $i\sim j$ if and only if $(i,j)$ lies in $G$, including the case $(i,i)=1$. Then $\sim$ is reflexive and symmetric, and we claim it is transitive. But if $(i,j)$ and $(j,k)$ are in $G$, they generate the subgroup $\mathrm{Sym}(\{i,j,k\})$ of $G$, which contains $(i,k)$. Hence $\sim$ is an equivalence relation. But then $G$ contains all transpositions of $S_n$, so is $S_n$.
Thus $G$ is simply a direct product of the symmetric groups on the orbits of $G$, and thus we can easily understand the intersection of any two such subgroups. $i$ and $j$ lie in the same orbit of $G\cap H$ if and only if they lie in the same orbit of $G$ and of $H$, and $G\cap H$ is the product of symmetric groups on those orbits (since $(i,j)$ lies in $G\cap H$).
The general case, even where $T$ consists of the set of elements of order $2$, is false, and counterexamples are numerous. One is given in the other answer to the question.
A: The more general statement is not true, even if you assume that $T$ is a set of elements of order $2$ like transpositions. For an explicit counter example, consider the dihedral group $D_8$ of order 16 which has presentation $$G=D_8=\langle s_1,s_2\mid s_1^2,\,s_2^2,\,(s_1s_2)^8\rangle.$$ Take as the set $T$ the set of all conjugates of $s_1$ and $s_2$ which all have order $2$. Thinking of $D_8$ as the group of symmetries of a regular octagon, the elements of $T$ are all of the reflectional symmetries.
Let $U_1$ be the subgroup of $G$ generated by $\{s_1,s_2s_1s_2\}\subset T$ which is isomorphic to the dihedral group $D_4$. Similarly, let $U_2$ be the subgroup of $G$ generated by $\{s_2,s_1s_2s_1\}\subset T$ which is also isomorphic to $D_4$. However you can check that $U_1\cap U_2$ is isomorphic to the cyclic group $\mathbb{Z}_4$ and consists of the rotations by multiples of $\pi/2$. Indeed $U_1\cap U_2\cap T=\emptyset$.

In fact this example is even closer to the case of $S_n$ than just the fact that $T$ consists of order $2$ elements. Both $S_n$ and $D_n$ are examples of Coxeter groups - groups generated by reflections. The set of transpositions in $S_n$ are the reflections in that case, and the set $T$ in my example above is the set of reflections for $D_n$.
Answer to the main question:
In this broader setting of Coxeter groups (or more accurately Coxeter systems) it is possible to justify the first statement in the case that $T$ is the set of transpositions of the form $(i\;i+1)$. With this particular choice of $T$, the pair $(S_n, T)$ is a Coxeter system. Writing $t_i=(i\;i+1)$ for $1\le i\le n-1$ then $S_n$ has the presentation $$\langle t_1,\dots, t_{n-1}\mid t_i^2\;\textrm{for}\; 1\le i\le n-1;\; (t_it_{i+1})^3\;\textrm{for}\; 1\le i\le n-2;\;\textrm{and} \;(t_it_j)^2\;\textrm{for}\; 1\le i\le n-1\;\textrm{such that}\;|i-j|\ge2\rangle.$$
(Compare the form of this presentation to the one above for $D_n$)
More generally a Coxeter system is a pair $(W,S)$ where $W$ is a group, $S=\{s_1,\dots,s_n\}\subset W$ generates $W$, and $W$ has a presentation of the form $$\langle s_1,\dots,s_n\mid s_i^2 \;\textrm{for}\;1\le i\le n;\;(s_is_j)^{m_{ij}}\;\textrm{for}\;1\le i\neq j\le n\rangle,$$ for suitable chosen integers $m_{ij}$.
It is a basic result in the theory of Coxeter groups that if $S'\subset S$ then $\langle S'\rangle$ is a Coxeter group with Coxeter system $(\langle S'\rangle,S')$.
Applying that in the case of $(S_n,T)$ then there are subsets $T_1,T_2\subset T$ such that $U_i=\langle T_i\rangle$, and then $U=U_1\cap U_2=\langle T_1\cap T_2\rangle$ is generated by $U\cap T$.
For an introduction to Coxeter groups see this set of notes, and the general theorem I apply is given in Theorem 5.2 there.
