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

Oct 2009
195
19


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.
 
Oct 2009
4,261
1,836


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
 
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Oct 2009
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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

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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.
 
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Oct 2009
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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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Oct 2009
195
19


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?
 
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