Indices and intersections

Suppose $A$ and $B$ are subgroups of a group $G$ , and suppose $B$ is of finite index in $G$ .

Show that the index of $A \cap B \leq A$ is finite, and in fact $| A : A \cap B | \leq | G : B |$ . Hint: Find a set map $A / A \cap B \to G / B$ .
Prove that equality holds in (a) if and only if $G = A B$ .

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Suppose $A$ and $B$ are subgroups of a group $G$, and suppose $B$ is of finite index in $G$.
\begin{enumerate}[topsep=0.1in]
	\item Show that the index of $A\cap B\leq A$ is finite, and in fact $|A:A\cap B|\leq |G:B|$. {\itshape Hint:} Find a set map $A/A\cap B\to G/B$.
	\item Prove that equality holds in (a) if and only if $G=AB$.
\end{enumerate}