Groups, show GxH is a group (final question)

[karnten07]

Homework Statement

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.

Homework Equations

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.

 

[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.

 

[karnten07]

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.

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?

 

[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?

 

[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.

 

[karnten07]

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.

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

Ive mentioend associativity now, thankyou for the tip

 

[morphism]

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

 

[karnten07]

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

The question gives the elements as a and b no?

 

[morphism]

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

 

[karnten07]

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

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

 

[HallsofIvy]

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

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?

 

[karnten07]

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?

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 doesn't mean they are in the group of G x H, so could i just use g and h?

 

[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

 

[HallsofIvy]

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 e_G is the identity in G, and e_H is the identity in H, what is the identity in GxH?

 

[ObsessiveMathsFreak]

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

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 existence 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.

 

[HallsofIvy]

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