Groups and Subgroups. Please advise!

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Chuckles

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Suppose that H and K are subgroups of G. Show that |HK| = |H|*|K| if and only if H∩K = {e}. Give an example.
 
My first instinct is to say that |HK| = |H|*|K| = e because of of group axiom 1, but that's a guess at best.

Honestly I have no idea how to go about this. I'm looking first for just a starting point, so any help you can give would be much appreciated.
 
Chuckles said:
My first instinct is to say that |HK| = |H|*|K| = e because of of group axiom 1, but that's a guess at best.

Honestly I have no idea how to go about this. I'm looking first for just a starting point, so any help you can give would be much appreciated.
Click to expand...

Saying |HK|=e makes no sense. |HK| is the number of elements in the set HK (this is not necessarily a group). Do you know what the set HK is? There is more than one way to do this problem, but it depends on what you are able to use.

A starting point might be to assume to that there is an \(\displaystyle e\neq a \in H\cap K\) (which IS a subgroup) and attempt to obtain a contradiction that \(\displaystyle |HK|=|H|*|K|\).
 
Chuckles said:
Suppose that H and K are subgroups of G. Show that |HK| = |H|*|K| if and only if H∩K = {e}. Give an example.
Click to expand...

I'm trying to do a bit of self-study on groups right now, so I'm not sure if this will help, but the way I interpret this is as follows:

If you have a set G, and you have both H and K as subgroups, that means that H and K are both groups themselves, but smaller factors of G which may or may not be "prime factors". If they only have {e} in common then they are both "prime factors" of G. By HK, I take that to mean "the image of H under K" and thus |HK| means "the order of H under K". Well, you started out with neither of them having anything in common (except {e}) and so the order of HK is the same as multiplying the order of H with the order of K.

If they only have {e} in common, then they are "prime factors" of G.


It's sometimes easier to put things in more familiar terms, so I'll do that here pretending that we're talking about real numbers, and substitute 2 for H and 3 for K and 6 for G.

When you look at the prime factors of H {1, 2} and K {1, 3}, the only number they have in common is one H∩K = {1}. If this is true, they are both prime numbers themselves, and so they have no common factors themselves, except the number one. Thus H has only two factors, 1 and H itself. Likewise K has only two factors, 1 and K itself. So H∩K = {1}

So, in a very crude way, |HK| = |H|*|K| if and only if H∩K = {e}
is comparable to saying:

The number of all the factors in 6 {1, 2, 3, 6} is the same as the number of factors in 2 {1, 2} multiplied by the number of factors in 3 {1, 3} if and only if 2 and 3 are prime numbers .

This so-called proof needs a lot of work, but I'm trying to simplify it at the expense of mathematical rigor.
 
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I think primality is too strong of an analogy. Relative primality would make your analogy better but still not perfect.

Example: H=Z/2Z x {0} and K= {0} x Z/2Z are subgroups of trivial intersection in G=Z/2Z x Z/2Z, and |G|=4=2*2=|H|*|K|
 
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