# Thread: [G : H],[G : K] finite -> [G:H n K] finite

1. ## [G : H],[G : K] finite -> [G:H n K] finite

I'm not looking for the solution, but a helpful first step.

So far, I realize that if $H \cap K = \{e\}$, then $[G : H \cap K$] is infinite. More generally, if $o(H \cap K) < \infty$, then $[G : H \cap K]$ is infinite.

Problem is, I don't see how to show anything more than $e \in H \cap K$

2. Originally Posted by davismj

I'm not looking for the solution, but a helpful first step.

So far, I realize that if $H \cap K = \{e\}$, then $[G : H \cap K$] is infinite. More generally, if $o(H \cap K) < \infty$, then $[G : H \cap K]$ is infinite.

Problem is, I don't see how to show anything more than $e \in H \cap K$

Let us denote by $G_N$ the set of left cosets of $G$ wrt some subgroup $N$ .

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

Tonio

3. Originally Posted by tonio
Let us denote by $G_N$ the set of left cosets of $G$ wrt some subgroup $N$ .

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

Tonio
If phi is well-defined and one-to-one, then $o(G_{H\cap K}) = o(G_H)o(G_K) \implies o(G_{H\cap K}) < \infty$?

4. Originally Posted by davismj
If phi is well-defined and one-to-one, then $o(G_{H\cap K}) = o(G_H)o(G_K) \implies o(G_{H\cap K}) < \infty$?
Yes, but be careful. These sets of cosets are...well....sets. In general the notation for order implies there is a group structure on this. This doesn't change the result, just a remark.

5. Originally Posted by tonio
Let us denote by $G_N$ the set of left cosets of $G$ wrt some subgroup $N$ .

Define now $\phi:\,G_{H\cap K}\rightarrow G_H\times G_K$ by $\phi(x(H\cap K)):=(xH,xK)$ . 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, $H \cap K = \{0\}$.

Thus $\phi (x)$ 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 $H \cap K$, this function is one-to-one. I don't see how you could make this claim if you don't know anything about $H \cap K$?

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

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

6. After looking at this for a little while longer, I realized that maybe I'm overcomplicating this; maybe I see the logic about why you don't really need to know the intersection of H and K to prove anything.