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

• karnten07
In summary: 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.
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.

## 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.

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.

morphism said:
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?

(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:
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.

morphism said:
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

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

morphism said:
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?

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

morphism said:
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

karnten07 said:
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?

HallsofIvy said:
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?

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

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?

NateTG said:
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.

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

## 1. What is the definition of a group?

A group is a mathematical structure that consists of a set of elements and a binary operation that combines any two elements in the set to produce another element in the set. The operation must also satisfy four properties: closure, associativity, identity, and inverse.

## 2. How do you prove that GxH is a group?

To show that GxH is a group, we must first show that the set GxH is closed under the binary operation. This means that for any elements (g1, h1) and (g2, h2) in GxH, their combination (g1g2, h1h2) is also in GxH. Next, we must show that the operation is associative, meaning that (g1g2)g3 = g1(g2g3) for any elements g1, g2, g3 in G and h1, h2, h3 in H. Then, we must find an identity element, which is the element (e, e) where e is the identity element in both G and H. Finally, we must show that every element in GxH has an inverse, meaning that for any (g, h) in GxH, there exists an element (g', h') in GxH such that (g, h)(g', h') = (e, e).

## 3. What is the significance of the direct product GxH?

The direct product GxH is significant because it allows us to construct a new group from two existing groups G and H. The elements of GxH are ordered pairs (g, h), where g is an element of G and h is an element of H. The operation on GxH is defined as (g1, h1)(g2, h2) = (g1g2, h1h2). This operation preserves the group properties of G and H, and the resulting group GxH has a larger set of elements and a more complex structure than either G or H alone.

## 4. Can GxH be a group if G and H are not groups themselves?

No, GxH can only be a group if both G and H are groups. The binary operation in GxH is defined in terms of the binary operation in G and H, so if G or H does not satisfy the group properties, the operation in GxH will not satisfy them either. Additionally, GxH must have an identity element and inverses for every element, which can only exist if G and H are groups.

## 5. What are some examples of groups that can be written as direct products?

Some examples of groups that can be written as direct products include the direct product of two cyclic groups, the direct product of two symmetric groups, and the direct product of a group and its center. Other examples include the direct product of the additive group of integers with the multiplicative group of positive rational numbers, and the direct product of two dihedral groups.

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