Direct Product of Groups: Subgroup Realization and Diagonal Subgroup

[Bleys]

I was reading on wikipedia on direct product of groups because I wanted find out if every subgroup of G \times H is realized as a direct product of subgroups of G and H. Apparently it is not, because the diagonal subgroup in G \times G disproves this. I'm a little confused, because I thought the proof I wrote was correct
for a subgroup write A \times B where A is a subset of G, and B a subset of H. Can't you show A is a subgroup of G using (g,1) and analogously with B? For example
m,n in A then (m,1),(n,1) are in A \times B. Hence (mn,1) is and therefore mn is in A?

There must be something wrong? Is the property true for certain type of groups? But I didn't use anything about G and H.

 

[M Quack]

Looks like you have shown that the product of any 2 subgroup A and B, A x B is a subgroup of the product group G x H.

You have not shown the opposite, that each subgroup of G x H can be written as a product A x B. Because that is not the case, as the counter example shows.

 

[DonAntonio]

I was reading on wikipedia on direct product of groups because I wanted find out if every subgroup of G \times H is realized as a direct product of subgroups of G and H. Apparently it is not, because the diagonal subgroup in G \times G disproves this. I'm a little confused, because I thought the proof I wrote was correct
for a subgroup write A \times B where A is a subset of G, and B a subset of H.

Here is the gist of this stuff: you can't do this. It is not true that any subgroup of the direct product \,G\times H\, can be realized as a subset of the corresponding cartesian product, just as it is not true that any subset of a cartesian product is a cartesian product of subsets of the corresponding sets in the product...

DonAntonio

Can't you show A is a subgroup of G using (g,1) and analogously with B? For example
m,n in A then (m,1),(n,1) are in A \times B. Hence (mn,1) is and therefore mn is in A?

There must be something wrong? Is the property true for certain type of groups? But I didn't use anything about G and H.

 

[Bleys]

Thanks for your replies!

It is not true that any subgroup of the direct product G×H can be realized as a subset of the corresponding
cartesian product, just as it is not true that any subset of a cartesian product is a cartesian product of subsets of the corresponding sets in the product...

I'm sorry I'm having a little trouble understanding. Isn't the cartesian product defined as the set of elements of the form (g,h). Then any subset is a set of this form as well, so it is another direct product? If it is, why aren't the summands subsets of their respective supersets?

 

[DonAntonio]

Thanks for your replies!



I'm sorry I'm having a little trouble understanding. Isn't the cartesian product defined as the set of elements of the form (g,h). Then any subset is a set of this form as well, so it is another direct product? If it is, why aren't the summands subsets of their respective supersets?

Very simple: take the set \,A:=\{1,2\}\,\text{ and its cartesian product}\,\,A\times A\, , and look at the latter's diagonal subset \,D:=\{(1,1)\,,\,(2,2)\}\,.

Well, try to represent \,D=X\times Y\,\,,\text{for some subsets}\,X,Y\subset A\, (Hint: you can't).

So, again, your claim in " Isn't the cartesian product defined as the set of elements of the

form (g,h). Then any subset is a set of this form as well" is false.

DonAntonio

 

[Bleys]

Ah, I forgot: the direct product includes all combinations of elements of the summands!
I also kept thinking the diagonal subset was some kind of pathological example (with B=A), but of course this works for general sets.

Thank you for explaining DonAntonio! :D