Direct Product of Groups

1. Jun 17, 2012

Bleys

I was reading on wikipedia on direct product of groups because I wanted find out if every subgroup of $G \times H$ is realised 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.

2. Jun 17, 2012

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.

3. Jun 17, 2012

DonAntonio

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

4. Jun 18, 2012

Bleys

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?

5. Jun 18, 2012

DonAntonio

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

6. Jun 18, 2012

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