Two group theory questions I have two questions here about this topic that have absolutely baffled me. Assistance in solving these would be appreciated.
Q1:
Let $S := \{(123),(235)\} \subseteq \Sigma_5$. Determine the subgroup $\langle$$S$$\rangle$ $\leq$ $\Sigma_5$.
Q2:
Let $G$ be a finite group with subgroups $A,B \leq G$. Show that
$$[G : A\cap B] = [G : A] \centerdot [A : A\cap B] \leq [G : A] \centerdot [G : B].$$

For question 1: As has been noted, since $4$ is fixed by both generators of $G$, we can imagine that we are actually working in $S_4$ (acting on ${1,2,3,5}$). Moreover, since $3$-cycles are even, we are actually inside of $A_4$; $A_4$ has $12$ elements, and $S$ contains at least $5$ elements (namely, $\{1\}$, $(123)$, $(132)$, $(235)$, and $(253)$), it must either have $6$ elements  or must be all of $A_4$ (by Lagrange's Theorem). However, $A_4$ is famous for being the smallest counterexample to the converse of Lagrange's Theorem: $A_4$ does not have subgroups of order $6$. Thus, $S$ must be all of $A_4$ (or rather, the even permutations in $S_5$ that fix $4$). If you want to explicitly produce the other seven elements explicitly in terms of the generators, we have (I compose my permutations right to left):
$$\begin{align*}
(123)(235) &= (12)(35)\\
(235)(123) &= (13)(25)\\
(123)(253) &= (125)\\
(253)(123) &= (153)\\
(132)(235) &= (135)\\
(235)(132) &= (152)\\
(152)(12)(35)(125) &= (15)(23)
\end{align*}$$
which gives all twelve elements of $A_4$.
For Question 2: Both the equality and the inequality hold for infinite groups.
Theorem. Let $K\leq H\leq G$ be groups; then $$[G:K]=[G:H][H:K]$$ in the sense of cardinalities.
Proof. Let $\{h_i\}_{i\in I}$ be a set of left coset representatives of $K$ in $H$; and let $\{g_j\}_{j\in J}$ be a set of left coset representatives for $H$ in $G$. We claim that $\{g_jh_i\}_{(i,j)\in I\times J}$ is a set of left coset representatives for $K$ in $G$.
Indeed, let $g\in G$. Then there exists $j\in J$ such that $g\in g_jH$; let $h\in H$ be such that $g=g_jh$. Then there exists $i\in I$ such that $h\in h_iK$. Therefore, there exists $k\in K$ such that $h=h_ik$. Hence, $g=g_jh_ik\in g_jh_i K$. That is: any element of $G$ is equivalent, modulo $K$ on the right, to some $g_jh_i$. 
Now assume that $g_jh_iK = g_rh_sK$; we need to show that $j=r$ and $i=s$. Indeed, there exists $k\in K$ such that $g_jh_i = g_rh_sk$. Since $k\in K\subseteq H$, then $g_jh_i\in g_jH$, and $g_rh_sk\in g_rH$. Hence, $g_jH\cap g_rH\neq\varnothing$, hence $j=r$ (since the $g_j$ are a set of coset representatives). Thus, $g_j=g_r$, so $g_jh_iK=g_rh_sK$ implies $h_iK = h_sK$; since the $h_i$ form a set of coset representatives for $K$ in $H$, it follows that $i=s$, as desired. $\Box$
Theorem. Let $G$ be a group, and let $A$ and $B$ be subgroups. Then
$$[A:A\cap B]\leq [G:B]$$
in the sense of cardinalities.
Proof. Let $\{a_i\}_{i\in I}$ be a set of coset representatives for $A\cap B$ in $A$. I claim that if $a_iB = a_jB$, then $i=j$. This will show that the number of left cosets of $B$ in $G$ is at least as large as $[A:A\cap B]$. 
Assume that $a_iB=a_jB$. Then there exists $b\in B$ such that $a_i = a_jb$. Hence $b=a_ia_j^{-1}\in A\cap B$, so $a_i\in a_j(A\cap B)$; hence, $i=j$, as desired. $\Box$
