Direct product of a subset of a group

[snipez90]

Homework Statement

http://math.uchicago.edu/~dc/teaching/254/254_problem_set_05.pdf"

Homework Equations

H a subgroup of G implies HH = H, where HH is the direct product.

The Attempt at a Solution

For the forward direction, don't we just have |A| = 0 or 1? I think |A| = 0 should be excluded since then A is the empty set, which makes proving the forward direction kind of nonsensical.

For the reverse direction, if A = aH = Hb, then AA = aHHb = aHb, and can't we just send each ah in aH (= A) to ahb in aHb (= AA) for the bijection?

This seems too easy, so I've likely made an oversight. Thanks in advance.

 

[micromass]

Your reverse implication is correct. But the forward implication is not.

If |A²|=|A|, then it isn't necessairily true that |A|=0 or 1. Take the entire group G for example, that satisfies |G²|=|G|, but it isn't, in general, true that |G|=0 or 1. No, you still have to do some work on the implication.

 

[snipez90]

Ah, I thought the forward direction was easier because |A^2| = |A|^2, a property that generalizes to cartesian products in general via a simple counting argument. But if this is the incorrect way to approach it, can you provide further suggestions? Thanks.

 

[micromass]

Maybe I'll give you something to get you started.

Consider x\in A. Then xA\subseteq A^2. But since |A|=|A^2|, we actually have xA=A^2.

Now the only thing you have to show is that x^{-1}A is a subgroup of G.

 

[micromass]

No, it is incorrect that |A²|=|A|²

Try to prove first that x^{-1}A is preserved under inverses. That's the hardest part I think...

 

[snipez90]

Thanks, I can certainly take it from there. I guess I'm just having a hard time convincing myself |A^2| =/= |A|^2. I've mostly been working with products of cyclic groups (e.g. in the fundamental theorem of finitely generated abelian groups), and I know the property certainly holds for such products, so I guess I'll look at other groups.

Blah, ok yeah I guess I should have looked at my relevant equation a bit longer.

 

[micromass]

But the property doesn't hold for cyclic groups... Let C2 be the cyclic group of order 2, then

|C_2|^2=4~\text{and}~|C_2^2|=2

 

[snipez90]

But C_2 is isomorphic to Z_2, and (Z_2)^2 is isomorphic to the Klein 4-group (often denoted V), which has 4 elements?

 

[micromass]

Yes, but you are confused. The A² here means something else, it means AA, not AxA...

 

[snipez90]

Ah dammit, sorry for the waste of time. The rest of the homework was on direct products, so that threw me off argh. I'll attempt to forget your hint for now and return to this in a day or two. Thanks.