# Groups, show GxH is a group (final question)

1. Feb 11, 2008

### karnten07

1. The problem statement, all variables and given/known data
Let G and H be groups. We define a binary operation on the cartesian product G x H by:

(a,b)*(a',b') := (a*a', b*b') (for a,a' $$\in$$G and b,b'$$\in$$)H

Show that G x H together with this operation is a group.

2. Relevant equations

3. The attempt at a solution
To be a group it must be a monoid (has an identity element) where every element has an inverse element.

I don't understand the binary operation, the commas confuse me. Any help would be most appreciated to complete this final question on this assignment.

2. Feb 11, 2008

### morphism

What don't you understand about the binary operation? The elements of GxH are those of the form (g,h) where g is in G and h is in H. As such, the binary operation defined just uses the binary operations of G and H coordinate-wise.

3. Feb 11, 2008

### karnten07

Oh yes i think i understand it better now. So the binary operation is just a multiplication and it shows this between an element and it's inverse.

To show each element of G x H has an identity element, could i just state, (a,b)*(a-1,b-1) = e

This also shows that there is an inverse i guess.

Im sure i'm missing the big picture here though, any more help please?

4. Feb 11, 2008

### karnten07

(a*b)*(b$$^{}-1$$*a$$^{}-1$$) = a*(b*(b$$^{}-1$$*a$$^{}-1$$)) = a*((b*b$$^{}-1$$)*a$$^{}-1$$) = a*(e*a$$^{}-1$$)=a*a$$^{}-1$$=e$$\Rightarrow$$(a*b)$$^{}-1$$= b$$^{}-1$$*a$$^{}-1$$

So here i have showed the binary operation of the elements of the group shows that there exists an inverse for the operation and that there is an identity element. Is this right?

Last edited: Feb 11, 2008
5. Feb 11, 2008

### morphism

What is the identity element in GxH?

Like you guessed, the most natural candidate for the inverse of (a,b) is (a-1, b-1). So this in turn implies that the most natural candidate for the identity of GxH is...?

Also, don't forget associativity.

6. Feb 11, 2008

### karnten07

(a*b)*(b^-1*a^-1) = e

Ive mentioend associativity now, thankyou for the tip

7. Feb 11, 2008

### morphism

What is e? The elements of GxH are ordered pairs of elements of G and H!

8. Feb 11, 2008

### karnten07

The question gives the elements as a and b no?

9. Feb 11, 2008

### morphism

But these are sets of elements you're dealing with -- of which a and b are just arbitrary representatives.

10. Feb 11, 2008

### karnten07

No it says a and b are in the set of G and H

11. Feb 11, 2008

### HallsofIvy

Staff Emeritus
Why do you start a sentence with "No" and then agree with him? This is exactly what Morphims said: "these are sets of elements you're dealing with", the sets G and H, " -- of which a and b are just arbitrary representatives."

Now, how about answering the question that morphism asked twice: What is the identity element of the group GxH?

12. Feb 11, 2008

### karnten07

I'm sorry i must have misunderstood. I thought that identity element, e, of GxH was (a*b)*(a^-1*b^-1). The reason i use a and b is because the binary operation is described with a and b as they are of the sets G and H respectively, but if you say it is only representatitve that is my mistake. Would i use g and h instead? The identity element of G x H then becoming (g*h)*(g^-1*h^-1)?

When i answered morphism's last question it was out of haste as i had to rush off to a lecture and hand in the work, so i apologise if it sounded blunt. I am very grateful for your guys' help and i hope that i will start to understand linear algebra and groups better when i read a bit more about it.

Edit: oh i think i see now, a and b are in the sets of G and H but that doesnt mean they are in the group of G x H, so could i just use g and h?

13. Feb 11, 2008

### NateTG

You should be able to come up with a much clearer cleaner expression for the identity than:
$$\left(a,b\right) \left(a^{-1},b^{-1}\right)$$

In order to show that it's a group you need
Identity
Inverses
Associativity
Closure over multiplication

14. Feb 11, 2008

### HallsofIvy

Staff Emeritus
It is certainly true that (ab)(ab)-1 or x x-1 are equal to the group identity! Morphism and I are asking what the group identity is in this particular group. Every member of GxH is of the form (a, b) where a is from G and b is from H. Now, if eG is the identity in G, and eH is the identity in H, what is the identity in GxH?

15. Feb 11, 2008

### ObsessiveMathsFreak

Done as follows
$$g \in G$$
$$h \in H$$
$$f=(g,h) \in G\timesH$$

For the proof of closure under multiplication.
$$\forall f_1=(g_1,h_1),f_2=(g_2,h_2) \in G\timesH$$
$$g_1,g_2 \in G \Rightarrow g_1 g_2 \in G$$
$$h_1,h_2 \in H \Rightarrow h_1 h_2 \in H$$
$$\Rightarrow f_1 f_2 = (g_1 g_2, h_1 h_2) \in G\timesH$$
Therefore the set is closed under multiplication.

To prove that $$G\timesH$$ has an Identity.

Let $$e_g \in G$$ and $$e_h \in H$$, be the identity elements in $$G\timesH$$. Then $$e_f = (e_g,e_h) \in G\timesH$$. We now show that $$e_f$$ is the identity.

$$\forall f \in G\timesH$$
$$e_f f = (e_g g, e_h h) = (g,h) = f$$
$$f e_f = ( g e_g , h e_h) = (g,h) = f$$
So
$$\forall f \in G\timesH , e_f f = f e_f = f$$
The existance of the identity element is proven.

For the prove of Inverses
$$f=(g,h) \in G\timesH$$
$$g \in G \Rightarrow g^{-1} \in G$$
$$h \in H \Rightarrow h^{-1} \in H$$
$$\Rightarrow (g^{-1},h{-1})=f^{-1} \in G\timesH$$ We now show that $$f^{-1}$$ is the inverse of $$f$$.

$$f^{-1} f = (g^{-1} g, h^{-1} h) = (e_g,e_f) = e_f$$
$$f f^{-1} = ( g g^{-1}, h h^{-1}) = (e_g,e_f) = e_f$$
So
$$\forall f \in G\timesH , f^{-1} f = f f^{-1} = e_f$$
So $$\forall f \in G\timesH$$ the inverse element exists.

Finally, to prove associativity.
$$\forall f_1=(g_1,h_1),f_2=(g_2,h_2),f_3=(g_3,h_3) \in G\timesH$$
$$f_1 (f_2 f_3) = ( g_1 (g_2 g_3 ), h_1 (h_2 h_3) )= (g_1 g_2 g_3, h_1 h_2 h_3 ) = ((g_1 g_2) g_3 , (h_1 h_2 ) h_3 ) = (f_1 f_2) f_3$$
$$\Rightarrow \forall F_1,f_2,f_3 \in G\timesH, , f_1 (f_2 f_3)=(f_1 f_2) f_3 = f_1 f_2 f_3$$

And this proves associativity. Hopefully that's all OK.

The prove that $$G\timesH$$ is a group is somewhat long but straightforward. It's a good exercise in the basic properties and laws of groups.

16. Feb 11, 2008

### HallsofIvy

Staff Emeritus
Well, I guess there is no point in trying to help karten07 work it out for himself any more!