Let us denote by \(\displaystyle G_N\) the set of left cosets of \(\displaystyle G\) wrt some subgroup \(\displaystyle N\) .

Define now \(\displaystyle \phi:\,G_{H\cap K}\rightarrow G_H\times G_K\) by \(\displaystyle \phi(x(H\cap K)):=(xH,xK)\) . Now show this map is a well-defined 1-1 function so...

Tonio

Okay. I'm still not getting it. Maybe this example is too abstract.

Lets consider the integers mod 12. Let H = {0,3,6,9} and K = {0,4,8}. Thus, \(\displaystyle H \cap K = \{0\}\).

Thus \(\displaystyle \phi (x)\) would be defined as

0 = ({0,3,6,9},{0,4,8})

1 = ({1,4,7,10},{1,5,9})

2 = ({2,5,8,11},{2,6,10})

3 = ({0,3,6,9},{3,7,11})

4 = ({1,4,7,10},{0,4,8})

5 = ({2,5,8,11},{1,5,9})

6 = ({0,3,6,9},{2,6,10})

7 = ({1,4,7,10},{3,7,11})

8 = ({2,5,8,11},{0,4,8})

9 = ({0,3,6,9},{1,5,9})

10 = ({1,4,7,10},{2,6,10})

11 = ({2,5,8,11},{3,7,11})

Now, the claim is that for any \(\displaystyle H \cap K\), this function is one-to-one. I don't see how you could make this claim if you don't know anything about \(\displaystyle H \cap K\)?

I notice, though, that in the above example, \(\displaystyle \phi(x)\) returns the cosets \(\displaystyle xH, xK \implies x \in xH \cap xK\). If this function is indeed one-to-one, there is one and only one \(\displaystyle g \in G\) with \(\displaystyle x \in gH \cap gK = g(H \cap K)\). Therefore, since there are finitely many such cosets, there are finitely many such g with \(\displaystyle \cup_{i \in I}g_i(H \cap K) = G\).

I guess I understand where this is going, but I don't know how to prove the injectivity if you don't know \(\displaystyle H \cap K\).