MHB What Do the Elements of $\mathbb{Z}_2 \times \mathbb{Z}$ Look Like?

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Elements
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

To show that in the additive group $\mathbb{Z}_2\times\mathbb{Z}$ there are non-zero elements $A,B$ of infinite order such that $A+B$ has finite order, we have to find such $A$ and $B$, right? (Wondering)

How do the elements of $\mathbb{Z}_2\times\mathbb{Z}$ look like? (Wondering)
 
Physics news on Phys.org
The elements of $\mathbb{Z}_2\times\mathbb{Z}$ are of the form $(x,y)$ where $x\in \mathbb{Z}_2$ and $y\in \mathbb{Z}$, right? (Wondering)

And the order of such an element is $n$ for which $(nx,ny)=(0,0)$, right?

For $A=(1,k)$ and $B=(0,-k)$ for $k\in \mathbb{Z}_{>0}$, we have that $A+B=(1,0)$, right?

$A$ and $B$ have infinite order, since the second coordinate is never equal to $0$ for $k>0$, and $A+B$ has finite order, and specifically the order is $2$, since $2(1,0)=(2,0)=(0,0)$.

Is this correct? (Wondering)
 
Yep. All correct. (Happy)
 
Great! Thank you! (Sun)
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
12K
Replies
17
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 26 ·
Replies
26
Views
708
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
7K
Replies
6
Views
2K