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

#### davismj 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 $$\displaystyle e \in H \cap K$$

#### tonio 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 $$\displaystyle e \in H \cap K$$

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

• davismj

#### davismj

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 $$\displaystyle o(G_{H\cap K}) = o(G_H)o(G_K) \implies o(G_{H\cap K}) < \infty$$?

#### Drexel28

MHF Hall of Honor
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.

• davismj

#### davismj

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 $$\displaystyle H \cap K$$.

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#### davismj 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.