Is $\Bbb Z_2 \times \Bbb Z_2$ a cyclic group?

[onie mti]

in Z_3 x Z_4 find all elements of cyclic subgroups(<1,2) generated by (1,2)

this is just confusing me :(

 

[Nono713]

Do you know what a generator is? Do you understand the group structure of $\mathbb{Z}_3$, $\mathbb{Z}_4$, and $\mathbb{Z}_3 \times \mathbb{Z}_4$?

 

[onie mti]

Do you know what a generator is? Do you understand the group structure of $\mathbb{Z}_3$, $\mathbb{Z}_4$, and $\mathbb{Z}_3 \times \mathbb{Z}_4$?

i am confused when it is z_4 x z_3

 

[Nono713]

i am confused when it is z_4 x z_3

That is another group, defined as the product of the groups $\mathbb{Z}_3$ and $\mathbb{Z}_4$, as:

$$\mathbb{Z}_3 \times \mathbb{Z}_4 = \left \{ \left ( x, y \right ) \mid x \in \mathbb{Z}_3, y \in \mathbb{Z}_4 \right \}$$

With the usual group operations operating component-wise, that is, $(x, y) +_1 (u, v) = (x +_2 u, y +_3 v)$, where $+_1$ is the group operation in $\mathbb{Z}_3 \times \mathbb{Z}_4$ (the one we're trying to define), $+_2$ is the group operation in $\mathbb{Z}_3$, i.e. addition modulo 3, and $+_3$ is the group operation in $\mathbb{Z}_4$.

But it's still a group, which means it still has all the properties you would expect from a group. In this case, then, what do you get when you repeatedly add $(1, 2)$ to itself? Note the plus sign here is the group operation in $\mathbb{Z}_3 \times \mathbb{Z}_4$, so use the definition above.

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

This is because the group operation in $\mathbb{Z}_4$ (the second component) sends $2 + 2$ to $0$, being modulo $4$. Then you can keep going:

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

Again, because the group operation in $\mathbb{Z}_3$ (the first component) sends $2 + 1$ to $0$.

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

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

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

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

And at this point you're back to the generator itself, $(1, 2)$, so you know that you've obtained all the elements generated by $(1, 2)$, which are:

$$\{ (0, 0), (1, 2), (2, 0), (0, 2), (1, 0), (2, 2) \}$$

Furthermore, since you generated six distinct elements, it means that $(1, 2)$ has order $6$ in this group. Notice that since $0$ is the identity in both $\mathbb{Z}_3$ and $\mathbb{Z}_4$, $(0, 0)$ is the identity in $\mathbb{Z}_3 \times \mathbb{Z}_4$. There are theorems that let you work out cyclic subgroups without going through all of that, but they require understanding the group structure of these groups, so doing it "by hand" to begin with is probably not a bad thing.

Does this look familiar to you? Can you connect it to ideas you've seen before with other, simpler groups?

 

[onie mti]

That is another group, defined as the product of the groups $\mathbb{Z}_3$ and $\mathbb{Z}_4$, as:

$$\mathbb{Z}_3 \times \mathbb{Z}_4 = \left \{ \left ( x, y \right ) \mid x \in \mathbb{Z}_3, y \in \mathbb{Z}_4 \right \}$$

With the usual group operations operating component-wise, that is, $(x, y) +_1 (u, v) = (x +_2 u, y +_3 v)$, where $+_1$ is the group operation in $\mathbb{Z}_3 \times \mathbb{Z}_4$ (the one we're trying to define), $+_2$ is the group operation in $\mathbb{Z}_3$, i.e. addition modulo 3, and $+_3$ is the group operation in $\mathbb{Z}_4$.

But it's still a group, which means it still has all the properties you would expect from a group. In this case, then, what do you get when you repeatedly add $(1, 2)$ to itself? Note the plus sign here is the group operation in $\mathbb{Z}_3 \times \mathbb{Z}_4$, so use the definition above.

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

This is because the group operation in $\mathbb{Z}_4$ (the second component) sends $2 + 2$ to $0$, being modulo $4$. Then you can keep going:

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

Again, because the group operation in $\mathbb{Z}_3$ (the first component) sends $2 + 1$ to $0$.

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

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

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

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

And at this point you're back to the generator itself, $(1, 2)$, so you know that you've obtained all the elements generated by $(1, 2)$, which are:

$$\{ (0, 0), (1, 2), (2, 0), (0, 2), (1, 0), (2, 2) \}$$

Furthermore, since you generated six distinct elements, it means that $(1, 2)$ has order $6$ in this group. Notice that since $0$ is the identity in both $\mathbb{Z}_3$ and $\mathbb{Z}_4$, $(0, 0)$ is the identity in $\mathbb{Z}_3 \times \mathbb{Z}_4$. There are theorems that let you work out cyclic subgroups without going through all of that, but they require understanding the group structure of these groups, so doing it "by hand" to begin with is probably not a bad thing.

Does this look familiar to you? Can you connect it to ideas you've seen before with other, simpler groups?

oh my word i had no idea that is how you work it out :) how do i tell if z_2 X z_2 is not cyclic

 

[Deveno]

Re: cyclic subgroup

It turns out that there is an easy test for this:

$\Bbb Z_m \times \Bbb Z_n$ is cyclic iff gcd($m,n$) = 1.

The reason being, is that for any $(a,b) \in \Bbb Z_m \times \Bbb Z_n$ it is not hard to see that if $k = \text{lcm}(m,n)$ that:

$k(a,b) = (ka, kb)$ and that $m|k$ and $n|k$, so that:

$ka = 0\text{ (mod }m)$
$kb = 0\text{ (mod }n)$

thus any element has order no greater than $k = \text{lcm}(m,n)$

However, $\Bbb Z_m \times \Bbb Z_n$ is cyclic iff it has an element of order $mn$.

Since $k = \text{lcm}(m,n) = \dfrac{mn}{\text{gcd}(m,n)}$, we see that we must have $\text{gcd}(m,n) = 1$.

On the other hand if $m,n$ are co-prime, then $(1,1)$ is easily seen to have order $\text{lcm}(m,n) = mn$, so that $\Bbb Z_m \times \Bbb Z_n$ is cyclic.

So, we can tell that $\Bbb Z_2 \times \Bbb Z_2$ cannot be cyclic, since gcd(2,2) = 2 > 1.

Another way to see this, is that every element in $\Bbb Z_2 \times \Bbb Z_2$ satisfies:

$(a,b) + (a,b) = (0,0)$, so that it has NO elements of order 4.

***********

Of course $\Bbb Z_2 \times \Bbb Z_2$ is small enough to check it "by hand", we have just 4 elements:

(0,0),(1,0),(0,1) and (1,1). Clearly the identity (0,0) has order 1 (like it does in ANY group).

(1,0) + (1,0) = (1+1,0+0) = (0,0) since 1+1 = 0 in $\Bbb Z_2$.

Similarly, (0,1) + (0,1) = (0+0,1+1) = (0,0) and (1,1) + (1,1) = (1+1,1+1) = (0,0).

So we have 3 elements of order 2, and one of order 1, whereas a cyclic group of order 4 has 2 elements of order 4, and just a single element of order 2, and one element of order 1.

As a side note, I remark that one way of showing two groups of the same order are NOT isomorphic, is to count how many elements of order 2 we have in each group. This doesn't ALWAYS work, but often it is quite efficient.

 

[simo1]

Re: cyclic subgroup

thank you so much

It turns out that there is an easy test for this:

$\Bbb Z_m \times \Bbb Z_n$ is cyclic iff gcd($m,n$) = 1.

The reason being, is that for any $(a,b) \in \Bbb Z_m \times \Bbb Z_n$ it is not hard to see that if $k = \text{lcm}(m,n)$ that:

$k(a,b) = (ka, kb)$ and that $m|k$ and $n|k$, so that:

$ka = 0\text{ (mod }m)$
$kb = 0\text{ (mod }n)$

thus any element has order no greater than $k = \text{lcm}(m,n)$

However, $\Bbb Z_m \times \Bbb Z_n$ is cyclic iff it has an element of order $mn$.

Since $k = \text{lcm}(m,n) = \dfrac{mn}{\text{gcd}(m,n)}$, we see that we must have $\text{gcd}(m,n) = 1$.

On the other hand if $m,n$ are co-prime, then $(1,1)$ is easily seen to have order $\text{lcm}(m,n) = mn$, so that $\Bbb Z_m \times \Bbb Z_n$ is cyclic.

So, we can tell that $\Bbb Z_2 \times \Bbb Z_2$ cannot be cyclic, since gcd(2,2) = 2 > 1.

Another way to see this, is that every element in $\Bbb Z_2 \times \Bbb Z_2$ satisfies:

$(a,b) + (a,b) = (0,0)$, so that it has NO elements of order 4.

***********

Of course $\Bbb Z_2 \times \Bbb Z_2$ is small enough to check it "by hand", we have just 4 elements:

(0,0),(1,0),(0,1) and (1,1). Clearly the identity (0,0) has order 1 (like it does in ANY group).

(1,0) + (1,0) = (1+1,0+0) = (0,0) since 1+1 = 0 in $\Bbb Z_2$.

Similarly, (0,1) + (0,1) = (0+0,1+1) = (0,0) and (1,1) + (1,1) = (1+1,1+1) = (0,0).

So we have 3 elements of order 2, and one of order 1, whereas a cyclic group of order 4 has 2 elements of order 4, and just a single element of order 2, and one element of order 1.

As a side note, I remark that one way of showing two groups of the same order are NOT isomorphic, is to count how many elements of order 2 we have in each group. This doesn't ALWAYS work, but often it is quite efficient.