Conclude the number of elements in $H$ divides the number of elements in $G$. Let $G$ be a finite group and $H$ be a subgroup of $G$. Show that if $aH$ intersect $bH$ is not empty, then $aH$ and $bH$ contain an equal number of elements in $G$. Conclude the number of elements in $H$ divides the number of elements in $G$. 
 A: Let $a_1H, a_2H, \ldots, a_mH$ be the set of all distinct cosets.  Recall that any two distinct cosets are disjoint.  Then, since $|a_iH| = |a_jH|$ for all  $i, j$, we take $\sum_{k=1}^m|a_k H| = m|H| = |G|$.
A: Two cosets $aH$ and $bH$ intersect iff $a=bh$ for some $h\in H$.  Thus if they intersect they are equal. 
It follows that the cosets partition $G$.  Since they all have the same order,  $\vert H\vert\mid\vert G\vert$.   
A: Actually, a much stronger statement holds:  with $H$ a subgroup of the finite group $G$, we have
$aH \cap bH \ne \emptyset \Longleftrightarrow aH = bH, \tag 1$
for if
$k \in aH \cap bH, \tag 2$
we have 
$\exists h_1, h_2 \in H, \; ah_1 = k = bh_2; \tag 3$
but then
$a^{-1}b = h_1h_2^{-1} \in H, \tag 4$
which implies
$a^{-1}bH = H, \tag 5$
or
$aH = bH. \tag 6$
Clearly $aH$ and $bH$ are of the same cardinality if (6) binds.
In fact, $aH$ and $bH$ have the same number of elements whether or not (1) holds.  Consider the map
$\theta: aH \to bH, \; \theta(ah) = ba^{-1}(ah) = bh; \tag 7$
$\theta$ is clearly surjective, since (7) shows 
$\theta(ah) = bh \tag 8$
for any $bh \in bH$; it is also injective, since
$\theta(ah_1) = \theta(ah_2) \Longrightarrow bh_1 = bh_2 \Longrightarrow h_1 = h_2; \tag 9$
since $\theta$ is bijective, the number of elements in $aH$ and $bH$ is the same.  Now every $a \in G$ is in some $aH$, namely $aH$ itself (recall that $a = ae \in aH$, where $e$ is the identity element of $G$); since the $aH$ are either disjoint are identical (by (1)), $G$ is partitioned into the disjoint cosets of $H$, each with an equal number of elements; thus the cardinality of $G$ is a multiple of the cardinality of $H$.
