## Direct product of Groups

For the discussion of math. Duh.

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jacksmack
Posts: 83
Joined: Wed Mar 21, 2012 2:18 am UTC
Location: Italy

### Direct product of Groups

Hi,

I have this exercise and there are various parts, I write down here only the first part.
Without the comprehension of this part I can't do anything in the remaining parts.

G and H are two groups.
Let GxH = {(x,y) : x ∈ G and y ∈ H}
The operation GxH consists of multiplying corresponding components:
(x,y) ∙ (x',y') = (xx', yy')
If G and H are donoted additively, it is customary to also denote GxH additively:
(x,y) + (x',y') = (x+x', y+y')

Part 1:
Prove that GxH is a group by checking the three group axioms:
(G1)
(x1,y1)[(x2,y2)(x3,y3)] =
[(x1,y1)(x2,y2)](x3,y3) =
(G2)
Let eG be the identity element of G, and eH the identity element of H.
Find and check the identity element of GxH?
(G3)
For each (a,b) ∈ GxH, finf and check the inverse of (a,b).

Part2:
List the elements of Z2xZ3, and write its operation table.
(Note: There are six elements, each of which is an ordered pair. The notation is additive).

development:
The are some other parts, but I stop here. I want to comprehend these ones.
I have tried somethings like these, but without succcess..:

For part1:
(G1)
(x1, y1)[(x2, y2)(x3, y3)] =
(x1, y1)(x2x3, y2y3) =
(x1x2x3, y1y2y3)

[(x1, y1)(x2, y2)](x3, y3) =
(x1x2, y1y2)(x3, y3) =
(x1x2x3, y1y2y3)

(G2)
If I am right, We have a single identity element for each group so:
(x, eH) ∙ (x', eH) = (xx', eH)
(eG, y) ∙ (eG, y') = (eG, yy')
(xx', eH) ∙ (eG, yy') = (xx'eG, eHyy')

(G3)
I have denoted (a,b)' as the inverse of (a,b):
(a, b) ∙ (a, b)' = (xx'eG, eHyy')

For Part2:
Z2xZ3 = {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)}
+_0_1_2
0_0_1_2
1_1_2_3

please can you help me? many thanks!
Last edited by jacksmack on Wed Oct 17, 2012 11:10 am UTC, edited 1 time in total.

Talith
Proved the Goldbach Conjecture
Posts: 848
Joined: Sat Nov 29, 2008 1:28 am UTC
Location: Manchester - UK

### Re: Direct product of Groups

You did associativity fine - you've shown that multiplying elements in any order gives the same element of the product. The identity and inverse part you have got a little lost on. The whole point is that we have this new group, called L say, and it happens to be made up from the groups G and H in a nice simple way as their cartesian product. In order to prove that L is a group, we need to find an element eL in L which acts as an identity on the other elements of L. Given that L=GxH, every element in L can be written as a pair of elements (one from G, one from H) so you need to find some element gG in G, and some element gH in H such that eL=(gG,gH) has the property of the identity on L. What suitable choices of elements do you think would be a good idea?

Hopefully, if you can find the identity element in L, you shouldn't have trouble finding the inverse of some arbitrary element l=(g,h) in L.

For part 2, I think you need to revise what a multiplication table of a group is. Every element in the group should appear once in both the top row and leftmost column.

jacksmack
Posts: 83
Joined: Wed Mar 21, 2012 2:18 am UTC
Location: Italy

### Re: Direct product of Groups

Talith wrote:What suitable choices of elements do you think would be a good idea?

is it right for find the identity element?
(x,y)∙(gG, gH) = (x,y)
(gG, gH) = (x,y)-1(x,y)
(gG, gH) = (e,e)
eL = (e,e)
check:
(x,y)∙(e,e) = (xe,ye) = (x,y)
(e,e)∙(x,y) = (ex,ey) = (x,y)

but i'm unable to find the inverse doing this:
(x,y)∙(x,y)-1 = (e,e)
(x,y)∙(x-1,y-1) = (e,e)

Talith wrote:you need to revise what a multiplication table of a group is.

for sure. We have Z2xZ3 = Z6, and I think that the six elements i have found are right.
So i should have a table with six rows and six columns

Talith
Proved the Goldbach Conjecture
Posts: 848
Joined: Sat Nov 29, 2008 1:28 am UTC
Location: Manchester - UK

### Re: Direct product of Groups

jacksmack wrote:is it right for find the identity element?
(x,y)∙(gG, gH) = (x,y)
(gG, gH) = (x,y)-1(x,y)

How do you know that an inverse exists yet? You can't just multiply by an element that you don't yet know the existence of. I should point out that you got to the write answer, but in maths that's not good enough if the way you got there isn't logical.

I would start my proof something like this. Let z be any element of GxH. From the definition of GxH, There exists an element g in G and an element h in H such that z=(g,h). We wish to find an element l (i'm deliberately not calling it e so that you don't start doing manipulations with it) in GxH such that

lz=z=zl,
that is,
l(g,h)=(g,h)=(g,h)l. [Call this equation (1)]

Given that we want l to be in GxH, we may assume that there exist elements gG in G, and gH in H such that l=(gG, gH). In order for l to satisfy equation (1) then,....

see if you can finish off that proof. As a tip, please try and use words in your proof. A proof should be an English sentence that can be read as prose - and it also makes it easier for us to see what logical steps you're making.

jacksmack wrote:We have Z2xZ3 = Z6
Where did that come from? It's certainly not true that these groups are equal as sets, let alone as groups. They are isomorphic as groups but it's best if you ignore that fact as you have probably not proven it yet in class and even if you have, you're not allowed to use that to find the multiplication table of the group Z2xZ3.

jacksmack
Posts: 83
Joined: Wed Mar 21, 2012 2:18 am UTC
Location: Italy

### Re: Direct product of Groups

Talith wrote:multiplication table of the group Z2xZ3

i'm trying this one. it could be the right way:
(Note: There are six elements, each of which is an ordered pair. The notation is additive).

Talith wrote:In order for l to satisfy equation (1) then,....

when we multiply gG by g, we would should obtain g, and when we multiply gH by h, we would should obtain h again.
So using the multiplication operation described at the starting point, we should obtain:
(gG,gH)(g,h) = (g,h) = (g,h)(gG,gH)
(gGg, gHh)=(g,h)=(ggG, hgH)
it's correct?

Talith
Proved the Goldbach Conjecture
Posts: 848
Joined: Sat Nov 29, 2008 1:28 am UTC
Location: Manchester - UK

### Re: Direct product of Groups

jacksmack wrote:
Talith wrote:In order for l to satisfy equation (1) then,....

when we multiply gG by g, we would should obtain g, and when we multiply gH by h, we would should obtain h again.
So using the multiplication operation described at the starting point, we should obtain:
(gG,gH)(g,h) = (g,h) = (g,h)(gG,gH)
(gGg, gHh)=(g,h)=(ggG, hgH)
it's correct?

Yes and now you're just 1 small step away from determining what gG and gH are. Remember, we want them to be elements of the groups G and H respectively. What elements from those groups do you know which satisfy the equation. Seeing as this is the last step, maybe you should try and write out the whole proof of this part from the very start (without just copying what I wrote in my last post) and also then show that the set GxH has inverses too (it's very similar to showing the existence of an identity).

jacksmack wrote:
Talith wrote:multiplication table of the group Z2xZ3

i'm trying this one. it could be the right way:
(Note: There are six elements, each of which is an ordered pair. The notation is additive).

The notation you've chosen here is a bit off and it's making you not see the pattern that you should be seeing. You've almost got it though. The only problem is that you're using a mixture of additive and multiplicative notation to correspond to what should be just one kind of group operation.

Perhaps if I show you a different example, you might get the general idea.

Let G=(Z2,+) and let H=(Z2,+). Where Z2 is the set {0,1} with the group operation is addition modulo 2.
The group GxH then has elements {(0,0), (0,1), (1,0), (1,1)}. As this group is obviously commutative (you might want to try and prove that if G is abelian, and H is abelian, then the group GxH is abelian), I'll only give the list of multiplications in this group up to commutativity.

(0,0)+(0,0)=(0+0,0+0)=(0,0)
(0,0)+(0,1)=(0+0,0+1)=(0,1)
(0,0)+(1,0)=(0+1,0+0)=(1,0)
(0,0)+(1,1)=(0+1,0+1)=(1,1)
(1,0)+(1,0)=(1+1,0+0)=(0,0)
(1,0)+(0,1)=(1+0,0+1)=(1,1)
(1,0)+(1,1)=(1+1,0+1)=(0,1)
(0,1)+(0,1)=(0+0,1+1)=(0,0)
(0,1)+(1,1)=(0+1,1+1)=(1,0)
(1,1)+(1,1)=(1+1,1+1)=(0,0)

When you're writing this out, you might want to underline the elements coming from H so that you can distinguish between elements from G and elements from H.

My advice if you're still getting confused is to go back to the defintion. It might seem very general at first, but really, it's no different to how you probably learned to add vectors in euclidean space together (in fact, you can think of vector addition as the group operation on the group RxRx....xR where R is the group of real numbers under addition. Can you see how this is a group? - stick to the case of RxR, ie vector addition in the plane, if this isn't too obvious at first.)

jacksmack
Posts: 83
Joined: Wed Mar 21, 2012 2:18 am UTC
Location: Italy

### Re: Direct product of Groups

Talith wrote:Remember, we want them to be elements of the groups G and H respectively. What elements from those groups do you know which satisfy the equation. Seeing as this is the last step, maybe you should try and write out the whole proof of this part from the very start (without just copying what I wrote in my last post) and also then show that the set GxH has inverses too (it's very similar to showing the existence of an identity).

G2)
so the whole proof from the beginning:
Let z be any element in GxH.
From the definition of GxH exists an element x in G and y in H such that z=(x,y).
We want to find an element l in GxH such that:

lz = z = zl
l(x,y) = (x,y) = (x,y)l (1)

We assume the existance of eG in G, and of eH in H such that l=(eG,eH).

We want that l satisfies the equation (1) so, we have to have that when we multiply eG by x, we obtain x, and when we multiply eH by y, we obtain y.
Therefore using the operation of multiplication defined at the starting point we get:
(eG, eH) = (x,y) = (x,y)(eG, eH)
(eGx, eHy) = (x,y) = (xeG, yeH)
therefore the identity in GxH is l=(eG, eH).

G3)
Considering one of the axioms of group:
For every element a in G, there exists an element a-1 in G such that aa-1=e AND a-1a=e.

Let z=(a,b) in GxH we have to find the element z-1:

zz-1=e AND
z-1z=e

therefore:

(a,b)z-1 = (eG) (1)
z-1(a,b) = (eG) (2)

in few word we have to find a particular ordered couple in GxH that multiplicated by the ordered couple (a,b), we obtain the identity element l=(eG, eH) (that we have found before).

Let rename z-1=(z-1a, z-1b)

>in (1) we obtain:
(a, b)(z-1a, z-1b) = (eG, eH)
(az-1a, bz-1b) = (eG, eH)

in a better words:
-when we go to multiply a by z-1a we have to obtain eG;
-when we go to multiply b by z-1b we have to obtain eH.

therefore:
az-1a = eG implies z-1a = a-1eG
bz-1b = eH implies z-1b = b-1eH

Let's go to substitute these values in (1):

(a,b)(a-1eG, b-1eH) = (eG, eH)
(aa-1eG, bb-1eH) = (eG, eH)
(eG, eH) = (eG, eH)

>in (2) we obtain:
(z-1a, z-1b)(a, b) = (eG, eH)
(z-1aa, z-1bb) = (eG, eH)

in a better words:
-when we go to multiply z-1a by a we have to obtain eG;
-when we go to multiply z-1b by b we have to obtain eH.

therefore:
z-1aa = eG implies z-1a = eGa-1
z-1bb = eH implies z-1b = eHb-1

Let's go to substitute these values in (2):

(eGa-1, eHb-1)(a,b) = (eG, eH)
(eGa-1a, eHb-1b) = (eG, eH)
(eG, eH) = (eG, eH)

So in finally, the inverse of (a,b) is:
(a-1eG, b-1eH) = (eGa-1, eHb-1)

Posts: 108
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### Re: Direct product of Groups

jacksmack wrote:We want to find an element l in GxH such that:

lz = z = zl
Let z=(a,b) in GxH we have to find the element z-1:

zz-1=e AND
z-1z=e

Why do you have two different names for the identity element of GxH? That's kind of a confusing trick to pull on people who are trying to read your work.

jacksmack wrote:therefore:
az-1a = eG implies z-1a = a-1eG
bz-1b = eH implies z-1b = b-1eH

"Our life is frittered away by detail. Simplify, simplify, simplify! I say, let your affairs be as two or three, and not a hundred or a thousand;" - Henry David Thoreau (emphasis mine)

You were practically done here, if only you had simplified your terms just a step further. After all, if you know what z-1a is, and you know what z-1b is, then you know what z-1 is. (Though I would have preferred the nomenclature z-1G and z-1H over z-1a and z-1b to refer to the G part and H part of z-1, respectively.)

As an example of what I mean, you ended by saying:

jacksmack wrote:So in finally, the inverse of (a,b) is:
(a-1eG, b-1eH) = (eGa-1, eHb-1)

Would you, or would you not think I was daft if I were to say:

"So in finality, the inverse of 3 is:
1/3 * 1 = 1 * 1/3"

And just leave it at that without simplifying?
HAL9000 wrote:The Passover bunny walks on water! So he be an representation of Horus!

jacksmack
Posts: 83
Joined: Wed Mar 21, 2012 2:18 am UTC
Location: Italy

### Re: Direct product of Groups

So for this part:
List the elements of Z2xZ3, and write its operation table.
(NOTE: There are six elements, each of which is an ordered pair. The notation is additive.)

Then we have:
Z2 = {0,1}
Z3 = {0,1,2}
Z2xZ3 = {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)}

So following the definition of addition of the exercise:
(0,0)+(0,0) = (0+0, 0+0) = (0,0)
(0,0)+(0,1) = (0+0, 0+1) = (0,1)
(0,0)+(0,2) = (0+0, 0+2) = (0,2)
(0,0)+(1,0) = (0+1, 0+0) = (1,0)
(0,0)+(1,1) = (0+1, 0+1) = (1,1)
(0,0)+(1,2) = (0+1, 0+2) = (1,2)

(0,1)+(0,0) = (0+0, 1+0) = (0,1)
(0,1)+(0,1) = (0+0, 1+1) = (0,2)
(0,1)+(0,2) = (0+0, 1+2) = (0,3)
(0,1)+(1,0) = (0+1, 1+0) = (1,1)
(0,1)+(1,1) = (0+1, 1+1) = (1,2)
(0,1)+(1,2) = (0+1, 1+2) = (1,3)

(0,2)+(0,0) = (0+0, 2+0) = (0,2)
(0,2)+(0,1) = (0+0, 2+1) = (0,3)
(0,2)+(0,2) = (0+0, 2+2) = (0,4)
(0,2)+(1,0) = (0+1, 2+0) = (1,2)
(0,2)+(1,1) = (0+1, 2+1) = (1,3)
(0,2)+(1,2) = (0+1, 2+2) = (1,4)

(1,0)+(0,0) = (1+0, 0+0) = (1,0)
(1,0)+(0,1) = (1+0, 0+1) = (1,1)
(1,0)+(0,2) = (1+0, 0+2) = (1,2)
(1,0)+(1,0) = (1+1, 0+0) = (2,0)
(1,0)+(1,1) = (1+1, 0+1) = (2,1)
(1,0)+(1,2) = (1+1, 0+2) = (2,2)

(1,1)+(0,0) = (1+0, 1+0) = (1,1)
(1,1)+(0,1) = (1+0, 1+1) = (1,2)
(1,1)+(0,2) = (1+0, 1+2) = (1,3)
(1,1)+(1,0) = (1+1, 1+0) = (2,1)
(1,1)+(1,1) = (1+1, 1+1) = (2,2)
(1,1)+(1,2) = (1+1, 1+2) = (2,3)

(1,2)+(0,0) = (1+0, 2+0) = (1,2)
(1,2)+(0,1) = (1+0, 2+1) = (1,3)
(1,2)+(0,2) = (1+0, 2+2) = (1,4)
(1,2)+(1,0) = (1+1, 2+0) = (2,2)
(1,2)+(1,1) = (1+1, 2+1) = (2,3)
(1,2)+(1,2) = (1+1, 2+2) = (2,4)

So the right operation table would be this:

_+__|__(0,0)__(0,1)__(0,2)__(1,0)__(1,1)__(1,2)
(0,0)|__(0,0)__(0,1)__(0,2)__(1,0)__(1,1)__(1,2)
(0,1)|__(0,1)__(0,2)__(0,3)__(1,1)__(1,2)__(1,3)
(0,2)|__(0,2)__(0,3)__(0,4)__(1,2)__(1,3)__(1,4)
(1,0)|__(1,0)__(1,1)__(1,2)__(2,0)__(2,1)__(2,2)
(1,1)|__(1,1)__(1,2)__(1,3)__(2,1)__(2,2)__(2,3)
(1,2)|__(1,2)__(1,3)__(1,4)__(2,2)__(2,3)__(2,4)

END.

>Another part of the exercise states this:
If G and H are abelian, prove that GxH is abelian.

So we need to prove that GxH=HxG:
(x’,y’)(x,y) = (x,y)(x’,y’)
(x’x, y’y) = (xx’, yy’)

Since G and H are abelian we have.
x’x = xx’ AND y’y = yy’

x = x’-1xx1
x = x

x’ = xx’x-1
x’=x’

y = y’-1yy’
y = y

y’ = yy’y-1
y’=y’

So GxH is abelian.

>Yet another part states:
Suppose the groups G and H both have the following property:
“Every element of the group is its own inverse”.
Prove that GxH also has this property.

Supposing that G has that property, and x,x’ are in G, so in this case we must have:
xx’ = e implies x = x’-1 AND x’ = x-1

so if:
xx’ = e
xx’ = x’x’-1
x(x’x’-1) = (x’x’-1)x’-1
x = x’-1

in the same way:
xx’ = e
xx’ = xx-1
(x-1x) = (x-1x)x-1
x’ = x-1

applying all these stuff to GxH, if we have (x,y) in GxH we must prove (if I’m right…) that:
(x,y)(x’,y’) = e implies (x,y) = (x’,y’) AND (x’,y’) = (x,y) -1
So:
(x,y)(x’,y’) = e
(x,y)(x’,y’) = (x’,y’)(x’,y’)-1
(x,y)[(x’,y’)(x’,y’) -1] = [(x’,y’)(x’,y’) -1](x’,y’) -1
(x,y) = (x’,y’) -1

In the same way:
(x,y)(x’,y’) = e
(x,y)(x’,y’) = (x,y)(x,y)-1
[(x,y)-1(x,y)](x’,y’) = [(x,y) -1(x,y)](x,y)-1
(x’,y’) = (x,y)-1

is it right? thanks!

Talith
Proved the Goldbach Conjecture
Posts: 848
Joined: Sat Nov 29, 2008 1:28 am UTC
Location: Manchester - UK

### Re: Direct product of Groups

jacksmack wrote:So for this part:
Z2xZ3 = {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)}

(0,1)+(0,2) = (0+0, 1+2) = (0,3)
(0,1)+(1,2) = (0+1, 1+2) = (1,3)
(0,2)+(0,1) = (0+0, 2+1) = (0,3)
(0,2)+(0,2) = (0+0, 2+2) = (0,4)
(0,2)+(1,1) = (0+1, 2+1) = (1,3)
(0,2)+(1,2) = (0+1, 2+2) = (1,4)
(1,1)+(0,2) = (1+0, 1+2) = (1,3)
(1,1)+(1,2) = (1+1, 1+2) = (2,3)
(1,2)+(0,1) = (1+0, 2+1) = (1,3)
(1,2)+(0,2) = (1+0, 2+2) = (1,4)
(1,2)+(1,1) = (1+1, 2+1) = (2,3)
(1,2)+(1,2) = (1+1, 2+2) = (2,4)

So the right operation table would be this:

_+__|__(0,0)__(0,1)__(0,2)__(1,0)__(1,1)__(1,2)
(0,0)|__(0,0)__(0,1)__(0,2)__(1,0)__(1,1)__(1,2)
(0,1)|__(0,1)__(0,2)__(0,3)__(1,1)__(1,2)__(1,3)
(0,2)|__(0,2)__(0,3)__(0,4)__(1,2)__(1,3)__(1,4)
(1,0)|__(1,0)__(1,1)__(1,2)__(2,0)__(2,1)__(2,2)
(1,1)|__(1,1)__(1,2)__(1,3)__(2,1)__(2,2)__(2,3)
(1,2)|__(1,2)__(1,3)__(1,4)__(2,2)__(2,3)__(2,4)

I've picked out the bits that you need to think very carefully about in the first part of your post. You seem to be a bit lost in the second part too, but let's clear the first part up first.

jacksmack
Posts: 83
Joined: Wed Mar 21, 2012 2:18 am UTC
Location: Italy

### Re: Direct product of Groups

ah yes!
I think, the first component of the ordered couple would be in modulo 2, while the second one in modulo 3.
So the right way would be this:
Z2xZ3 = {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)}

(0,1)+(0,2) = (0+0, 1+2) = (0,0)
(0,1)+(1,2) = (0+1, 1+2) = (1,0)
(0,2)+(0,1) = (0+0, 2+1) = (0,0)
(0,2)+(0,2) = (0+0, 2+2) = (0,1)
(0,2)+(1,1) = (0+1, 2+1) = (1,0)
(0,2)+(1,2) = (0+1, 2+2) = (1,1)
(1,1)+(0,2) = (1+0, 1+2) = (1,0)
(1,1)+(1,2) = (1+1, 1+2) = (0,0)
(1,2)+(0,1) = (1+0, 2+1) = (1,0)
(1,2)+(0,2) = (1+0, 2+2) = (1,1)
(1,2)+(1,1) = (1+1, 2+1) = (0,0)
(1,2)+(1,2) = (1+1, 2+2) = (0,1)

So the right operation table would be this:

_+__|__(0,0)__(0,1)__(0,2)__(1,0)__(1,1)__(1,2)
(0,0)|__(0,0)__(0,1)__(0,2)__(1,0)__(1,1)__(1,2)
(0,1)|__(0,1)__(0,2)__(0,0)__(1,1)__(1,2)__(1,0)
(0,2)|__(0,2)__(0,0)__(0,1)__(1,2)__(1,0)__(1,1)
(1,0)|__(1,0)__(1,1)__(1,2)__(0,0)__(0,1)__(0,2)
(1,1)|__(1,1)__(1,2)__(1,0)__(0,1)__(0,2)__(0,0)
(1,2)|__(1,2)__(1,0)__(1,1)__(0,2)__(2,0)__(0,1)