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

• May 29th 2010, 04:47 PM
davismj
[G : H],[G : K] finite -> [G:H n K] finite
http://i48.tinypic.com/s46qyq.png

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

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

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

Thanks for your help.
• May 29th 2010, 06:15 PM
tonio
Quote:

Originally Posted by davismj
http://i48.tinypic.com/s46qyq.png

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

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

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

Thanks for your help.

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
• May 29th 2010, 07:13 PM
davismj
Quote:

Originally Posted by tonio
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

If phi is well-defined and one-to-one, then $\displaystyle o(G_{H\cap K}) = o(G_H)o(G_K) \implies o(G_{H\cap K}) < \infty$?
• May 29th 2010, 10:15 PM
Drexel28
Quote:

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

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.
• May 31st 2010, 11:02 AM
davismj
Quote:

Originally Posted by tonio
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 $\displaystyle H \cap K$.
• May 31st 2010, 12:09 PM
davismj
http://i50.tinypic.com/rcslcn.png

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.

Does this look about right?