# Thread: When is a Cartesian Product of Two Groups A Group?

1. ## When is a Cartesian Product of Two Groups A Group?

Hi Everyone,
My problem is stated as: "Determine the groups G such that $\{(g,g^{-1})\mid g \in G\}$ is a subgroup on $G\times G$." To me, appears that if G is a group, then this should always define a subgroup.

Going through the subgroup test:
Identity: All of the resulting subgroups will have an identity, namely $(1_G,1_G)$.
Units: All elements are invertible. For any pair $(g,g^{-1})$, the inverse is $(g^{-1},g)$, which also exists in the subgroup.
Closed: For any $a,b\in G$, we would have $(a,a^{-1})(b,b^{-1})=(ab,a^{-1}b^{-1})$, which should also exist in the subgroup.

Is there something I'm missing? The question seems to imply that in some cases, this subset will not be a subgroup, but I don't see when that would occur.

Thanks,
Peter

2. ## Re: When is a Cartesian Product of Two Groups A Group?

Hey flybynight.

What about associativity? Is there a condition with regards to (ab)^(-1) = b^(-1)*a^(-1)? Does this have implications on the associative axiom?

3. ## Re: When is a Cartesian Product of Two Groups A Group?

I could be wrong, but my thinking was that the subgroup inherits associativity from the parent group, $G\times G$, so that shouldn't play a role, should it? Or am I misunderstanding what you mean?

4. ## Re: When is a Cartesian Product of Two Groups A Group?

I was thinking something along the lines of (ab)^(-1)c^(-1) != a^(-1)(bc)^(-1) in general.

5. ## Re: When is a Cartesian Product of Two Groups A Group?

I promise I'm not being purposely dense. If I understand you correctly, we're dealing with something like: $(a,a^{-1})(b,b^{-1})(c,c^{-1})$, which, when isolating the second element of each pair, can be grouped in two separate ways: $(a^{-1}b^{-1})c^{-1}$ or $a^{-1}(b^{-1}c^{-1})$. But when doing the operation in G, there is no real difference between the operations, so far as I can tell.

What you're saying seems to be something like: consider $(ab,(ab)^{-1})(c,c^{-1})$ and $(a,a^{-1})(bc,(bc)^{-1})$, which isolating the second term, gives $(ab)^{-1}c^{-1}$ and $a^{-1}(bc)^{-1}$, respectively. However, the fact that these terms are different isn't problematic, so far as I can see. The answer will be different, but that makes sense, given that the starting elements are different. Unless, again, I'm confused.

6. ## Re: When is a Cartesian Product of Two Groups A Group?

I'm just giving some food for thought and not a solution: if my reasoning is wrong then like you, I would like to see where I have slipped up.

If the associativity fails then you don't have a group. So if (ab)^(-1)c^(-1) must equal a^(-1)(bc)^-1 then you don't have a group since a^(-1)(bc)^-1 = a^(-1)c^(-1)*b^(-1) and (ab)^(-1)c^(-1) = b^(-1)a^(-1)c^(-1) so you don't get an equality unless you impose very special conditions.

7. ## Re: When is a Cartesian Product of Two Groups A Group?

here is the problem:

it's NOT associativity:

[(g,g-1)*(h,h-1)]*(k,k-1) =

(gh,g-1h-1)*(k,k-1) =

((gh)k,(g-1h-1)k-1) =

= (g(hk),g-1(h-1k-1)) (by associativity of the group product in G)

= (g,g-1)*[(h,h-1)*(k,k-1)]

it's NOT identity:

(g,g-1)*(e,e) = (ge,g-1e) = (g,g-1) (note that e-1 = e)

(e,e)*(g,g-1) = (eg,eg-1) = (g,g-1)

it's NOT inverses:

(g-1,g) is clearly a two-sided inverse for (g,g-1).

hmm...well, then what is the problem?

the most BASIC fact of what a binary operation on a set MEANS: if a,b are in S, then a*b HAS TO BE IN S ALSO!

now (g,g-1)*(h,h-1) = (gh,g-1h-1)

is g-1h-1 equal to (gh)-1? what do we call the groups in which this holds true?

8. ## Re: When is a Cartesian Product of Two Groups A Group?

I'm not sure if I'm on the right track, but as far as I can see, if g-1h-1=(gh)-1, then the group is Abelian. Is this correct?

9. ## Re: When is a Cartesian Product of Two Groups A Group?

give this person a prize! how about vice-versa?

10. ## Re: When is a Cartesian Product of Two Groups A Group?

What do you mean, vice versa? If the group is Abelian, then the equivalence holds? Also, I don't know if I am completely clear on why it is important that g-1h-1=(gh)-1. If you have (g,g-1)*(h,h-1)=(gh,g-1h-1), g-1h-1 should be in G, because G must include both inverses, and because G is closed, it must also include their product, right?

I really do appreciate you helping me,
Peter

11. ## Re: When is a Cartesian Product of Two Groups A Group?

yes G is closed, however the subset of GxG consisting of pairs (g,g-1) may NOT be.

look, let's look at an example, so you can see why this fails:

let G = {e,a,a2,b,ab,a2b} where a3 = b2 = e, and ba = a2b.

the last rule: ba = a2b let's us "move a's to the front", so which know WHICH element of G any product is. for example:

(ab)(a2b) = a(ba)(ab) = a(a2b)(ab) = (a3b)(ab) = b(ab) = (ba)b = (a2b)b = a2b2 = a2.

i'll list the inverses for you (you can verify them if you like):

e-1 = e (duh!)
a-1 = a2 (since a3 = e)
(a2)-1 = a
b-1 = b
(ab)-1 = ab (we can show this one of two ways: (ab)-1 = b-1a-1 = ba2 = (ba)a = (a2b)a = a2(ba) = a2(a2b) = a4b = ab, or:

(ab)(ab) = a(ba)b = a(a2b)b = a3b2 = ee = e, so ab is of order 2 and thus its own inverse)

(a2b)-1 = a2b

now the subset of GxG we are considering is the set:

H = {(e,e),(a,a2),(a2,a),(b,b),(ab,ab),(a2b,a2b)}

let's look at:

(a,a2)*(ab,ab) = (a(ab),a2(ab)) = (a2b,a3b) = (a2b,b)

hmm...does it look like b is the inverse of a2b? is the pair (a2b,b) in H?

the closure of G only implies that the product of two elements (under * in GxG) is again in GxG. GxG has |G|2 elements. the subset:

H = {(g,h) in GxG: h = g-1} only has |G| elements, it's a MUCH smaller set.

12. ## Re: When is a Cartesian Product of Two Groups A Group?

Thank you! That clears it up!