Find the order of the group $G=\langle (1,2,3,4,5,6,7),(1,2,4)(3,6,5)\rangle$. Let $G=\langle (1,2,3,4,5,6,7),(1,2,4)(3,6,5)\rangle$. Find the order of $G$.
Here is a proof from class that I don't quite understand.

Set $k=(1,2,3,4,5,6,7),\,h=(1,2,4)(3,6,5),\,K=\langle k\rangle,\,H=\langle h\rangle$. First we claim that $G=HK$. Notice that $K=\langle k\rangle$ is the smallest subgroup containing $k$, and that $G$ contains $k$. Hence $K=\langle k \rangle$ lies in $G$ and similarly for $\langle h\rangle$. So the RHS lies fully in the LHS. We can show that $H$ normalises $K$ (omitted) and hence the product $HK$ is a subgroup. This subgroup contains $k$ and contains $h$, so the LHS being the smallest subgroup containing $h$ and $k$ must lie fully in the RHS. Thus $HK=G$ is a group and so $$|G|=|HK|=\frac{|H||K|}{|H\cap K|}=\frac{3\cdot 7}{1}=21.$$

Two questions:
1) Why is it true that both parts of the RHS lying fully in the LHS should imply that their product must also lie in the LHS?
2) When we say "lie in", I assume we mean as a subset rather than as a subgroup (since for example we said the RHS lies in the LHS without assuming the RHS is a subgroup) so why does it make sense to say $HK$ is a group? Isn't it only equal to the set $G$? Another similar question is that if we show a set $X\subseteq A_5$, say, with $|X|=60=|A_5|$ again why should this mean $X=A_5$ is a subgroup as opposed to just being equal to $A_5$ as a set?
 A: G is a Fano group.
The symbol (1,2,3,4,5,6,7), is a cyclic rotation of the seven positions.
The (1,2,4)(3,6,5) means that you rotate both sets in step, so
1,2,3,4,5,6,7  gives 2,4,6,1,3,5,7 gives 4,1,5,2,6,3,7.
If we rotate the first by 1 step, we get
2,3,4,5,6,7,1  giving 3,5,7,2,4,6,1 gives 5,2,6,3,7,4,1
So this group corresponds to the rotation and isomorphism of the heptagon.
There are 21 positions, so the answer is correct.
Question (1)  
I've expanded out the "cosets" of  against two members of .  Since  by itself is closed, and contains only one member of , there are at most gh elements.
Here, g and h are co-prime.  Therefore every member of this group must lie in a co-set (or set parallel to) the operation of  and the operation of .
Question (2)
In the group A5, you could have (1,2,3,4,5), (1,2,3).  This is the rotational symmetry of the dodecahedron, for example.  By itself, (1,2,3,4,5) could represent a rotation of the dodecahedron, with the axies running through the pole.  This is not the complete A5, so we suppose it's a subset.
But the two operators together suffice to produce all sixty even permutations.
Saying then that X is in Y, means that every operation of X is found in Y.  This could be the full set, or it could be a proper subset.  The usual test of equality is to show all Y is in X, whence X=Y.
