Answer: Formal Proof of G x H as a Group

[smoothman]

Let G = (G, . , e), H = (H , * , E) be groups ... (e is the identity)
the direct product is defined by:

G x H = (GXH, o , (e,E)) where,
(g1,h1) o (g2,h2) = (g1 . g2, h1*h2)

Question: Show formally that G x H is a group.
when it says, "show formally that G x H is a group".. does it mean show that G x H satisfies the 4 conditions for being a group.. i.e.

associativity,
closure,
existence of identity element and
existence of inverse element?

any help is appreciated..

 

[dodo]

It looks pretty straightforward. Basically you are being asked to do something like this:

1) If G is a group, and g1, g2 are elements of G, then <some property> applies to g1, g2.
2) If H is a group, and h1, h2 are elements of H, then <some property> applies to h1, h2.
3) Now, given the cartesian product G x H, and (g1,h1), (g2,h2) being elements of G x H,
use 1) and 2) to prove that <some property> applies to (g1,h1), (g2,h2).

Repeat four times, once per property to be proved.

. . . . . . .

That is, if I'm interpreting correctly the rather confusing usage of "G x H" and "GXH", as the same but not the same. Confusing.

 

[smoothman]

yes sorry, GXH = G x H...

i understand <some property> for 1 and 2 are "." and "*" respectively.. where these are the identity elements.
"o" applies to (3).

but what do u mean by "repeat four times".. repeat what 4 times>?

 

[dodo]

When you have vectors in the plane, they are represented by pairs of coordinates (x1,y1), (x2,y2), where the coordinates are real numbers. If you know how to add vectors, you can prove that the set of all vectors form a group with respect to vector addition; you use the known properties of the addition of real numbers to prove things about the addition of vectors.

Your problem is the same, only that you deal with 'vectors' of the form (a,x), (b,y), (c,z), where a,b,c are drawn from G, and x,y,z from H.

. . .

By "repeat four times", I meant that you need to prove 4 things: associativity, closure, existence of identity and existence of inverse.

 

[HallsofIvy]

yes sorry, GXH = G x H...

i understand <some property> for 1 and 2 are "." and "*" respectively.. where these are the identity elements.

You understand wrong. <some property> refers to "Associativity", "Closure", "Existence of an identity", and "existence of inverses". Exactly the properties you mentioned.

"o" applies to (3).

but what do u mean by "repeat four times".. repeat what 4 times>?

Prove that eaach of the four properties you mentioned hold.

 

[HallsofIvy]

When you have vectors in the plane, they are represented by pairs of coordinates (x1,y1), (x2,y2), where the coordinates are real numbers. If you know how to add vectors, you can prove that the set of all vectors form a group with respect to vector addition; you use the known properties of the addition of real numbers to prove things about the addition of vectors.

Your problem is the same, only that you deal with 'vectors' of the form (a,x), (b,y), (c,z), where a,b,c are drawn from G, and x,y,z from H.

Your "vector" analogy doesn't quite work- there is no "scalar multiplication" in a direct product of groups.

. . .

By "repeat four times", I meant that you need to prove 4 things: associativity, closure, existence of identity and existence of inverse.

 

[dodo]

Your "vector" analogy doesn't quite work- there is no "scalar multiplication" in a direct product of groups.

There was no intention to show that the direct product of groups forms a vector space. I was merely providing an illustration for the 'long and scary' term 'Cartesian product'. :)

 

[smoothman]

ok:
so closure is satisfied because we define a binary operation $*$ such that $*:G\times G\mapsto G$, so the group is closed.

existence of identity element is satisfied
because if we choose $(e,E)$ where $e$ is identity element in $G$ and $E$ is identity element in $H$. Then $(g,h)*(e,E) = (ge,hE)=(g,h)$. And similarly $(e,E)*(g,h) = (g,h)$.

question:
having problems proving associativity and existence of inverse element. any help is appreciated :)

 

[dodo]

Your work for the existence of identity was on the right direction; in fact, you would use the same technique for associativity or for the existence of inverse.

For example, for associativity, you know where to start:

$$\left( \ (g_1,h_1) \ \mbox{"o"} \ (g_2,h_2) \ \right) \ \mbox{"o"} \ (g_3,h_3)$$​

and where you want to arrive:

$$(g_1,h_1) \ \mbox{"o"} \ \left( \ (g_2,h_2) \ \mbox{"o"} \ (g_3,h_3) \ \right)$$​

where (g1,h1), (g2,h2) and (g3,h3) are elements of G x H.

You also know the definition of the "o" operation of G x H, in terms of the "dot" and "asterisk" operations of G and H resp. So you substitute this definition into your "starting point" equation. Notice how applying the definition of "o" is a way of "changing context" from the 'unknown world' of G x H to the 'known world' of G and H, a world where you know the group conditions to work.

Then, you try to apply the (known to exist) associativity in G and H, that is,

$$\left( \ g_1 \ . \ g_2 \ \right) \ . \ g_3 \ = \ g_1 \ . \ \left( \ g_2 \ . \ g_3 \ \right)$$​

and

$$\left( \ h_1 \ . \ h_2 \ \right) \ . \ h_3 \ = \ h_1 \ . \ \left( \ h_2 \ . \ h_3 \ \right)$$​

And then you use again the definition of "o" in terms of "dot" and "asterisk", to bring back your statement from the context of G and H to the context of G x H.

You did the same thing with the existence of identity: you used the definition of "o" to go from the operation (g,h) o (e,E) to the single pair (g . e, h $$*$$ E); then applied the (known to work) identities on each individual component.

. . .

By the way, be careful with the notation: "o" is the operation between pairs (g1,h1) and (g2,h2), while the dot is an operation between g's and the asterisk is an operation between h's. They are not interchangeable!

 

[smoothman]

i understand exactly what you mean and i will take care of the notation in future. just for the purposes of this question i have completed the associtivity and existence of inverse element using the same notation as before.. is this correct:



associativity:
$(g, h) * [(g', h') * (g'', h'')] <br /> =(g,h) * ((g'g'', h'h'')<br /> = (gg'g'', hh'h'') ..(1)$

now
$[(g, h) * (g', h')] * (g'', h'')<br /> = (gg', hh')*(g'', h'')<br /> = (gg'g'', hh'h'') ..(2)$

(1) & (2) are equal so associativity exists.

existence of inverse element:
$(g,h)*(g^{-1}, h^{-1}) = (gg^{-1}, hh^{-1}) = (1,1) = (e,E) .. (1)$
and
$(g^{-1}, h^{-1})*(g,h) = (g^{-1}g, h^{-1}h) = (1,1) = (e,E) .. (2)$

(1) & (2) are equal so the inverse element exists.

 

[mathwonk]

if you can prove the set of all functions from a set S into a group, is also a group, then this is almost the special case where the domain of the functions is two points.

 

[dodo]

Smoothman, I think your reasoning shows that you got the idea and are going in the right direction... but the notation is going to kill you. If you leave so much things "assumed and not said", depending on the teacher s/he would either think that you were lazy to write too much or that you don't know.

So let me attempt some precisions here.

• That "o", and not "asterisk", is the proper operation between pairs, we already talked about.

• On a similar tone, whenever you say $$(g g', h h')$$ you really mean $$(g \ \mbox{"dot"} \ g', h \ \mbox{"asterisk"} \ h')$$. Not writing any operator at all, could be interpreted by the teacher in the two ways I mentioned above; one is not quite good and the other is really bad for you. Take the control of your marks in your hands!

• This one is more important. If you try to prove the existence of inverses, you should not use inverses as if they existed. (They exist on G and on H, separately; so you can use $$g^{-1}$$ or $$h^{-1}$$ along your demonstration. But you are explicitly saying nowhere what is your candidate for the inverse of an element (g,h) of G x H. It's just a matter of words, but not saying "this proves that the inverse of (g,h) is ... whatever you found ...", could cost you the marks.

• Another little detail, but one that might lead the teacher to think that you don't know: There is no such thing here as (1,1). I see where you want to go when you say $$(g g^{-1}, h h^{-1}) = (1,1) = (e, E)$$, but the fact is that the identity for the g's is "e", the identity for the h's is E, and none of them is "one". There is no such thing as "one" here, since these are not numbers, just abstract things belonging resp. to G and H.

So, you seem to get the idea on how to solve this, but... is your answer right or not? Toss a coin. Or fix the notation.

. . .

I didn't mention this earlier in order to concentrate on the work, but regarding closure, I don't think closure can be taken from granted just like that, especially if you're being asked to prove "formally" that G x H is a group. Again, I can't make sense of this

$$*:G\times G\mapsto G$$

because of the notation: the "asterisk" is not the operation in G. Nor it is the one you are being asked to prove closed, that would be the "o", which doesn't go from G x G to G.

But you do know that the "dot" is closed in G, and that the "asterisk" is closed in H. So you follow the same procedure to prove that the "o" is closed in G x H. First, move from the 'unknown world' to the 'known world'; then apply the 'known' conditions. You want to prove that the result of $$(g,h) \ \mbox{"o"} \ (g', h')$$ is another element $$(g'',h'')$$ belonging to G x H.

 

[smoothman]

thnx dodo. i agree completely with what you said. on my final presentation, i have fixed all the mistakes in notation to make it clear for the teacher. i appreciate everything u said, which is correct :) thnx for the help.

 

[smoothman]

i also understand where i have gone wrong :) thnx