# Group elements order

1. Aug 15, 2012

### kljoki

Hi
i need a little help
i was given group (Z3 x Z3,+) and i should find order of every elements
so the elements are {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),( 2,2)} and the order of every element is
(0,0) has order 1
(0,1)*3=(0(mod 3),3(mod 3)) = (0,0) order 3
(0,2)*3=(0(mod 3),6(mod 3)) = (0,0) order 3
(1,0)*3=(3(mod 3),0(mod 3)) = (0,0) order 3
(1,1)*3=(3(mod 3),3(mod 3)) = (0,0) order 3
(1,2)*3=(3(mod 3),6(mod 3)) = (0,0) order 3
(2,0)*3=(6(mod 3),0(mod 3)) = (0,0) order 3
(2,1)*3=(6(mod 3),3(mod 3)) = (0,0) order 3
(2,2)*3=(6(mod 3),6(mod 3)) = (0,0) order 3

(a,b) + (a,b) + (a,b) = (3a(mod3), 3b(mod3))=(0,0) so max order is 3

next is group (Z2 x Z4, *)
the elements are {(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)}
so here i should multiply every element n times till i get (an(mod 2),bn(mod 4)) = (1,1) so the order is n (i'm not sure about this correct me if i'm wrong)
the element (0,0) always have order one
and what about the other elements???

example the element (0,2)
there isn't ANY n with (0,2)n (mod 2, mod 4) = (1,1)

2. Aug 15, 2012

### micromass

Staff Emeritus
But $\mathbb{Z}_2\times\mathbb{Z}_4$ isn't even a group under multiplication. Are you sure they didn't say that the operation is addition?

3. Aug 15, 2012

### christoff

On top of that, $\mathbb{Z}_4$ itself isn't a group under multiplication either. At best, it could be a field (if 4 were prime), so talking about a product like $(\mathbb{Z}_2\times\mathbb{Z}_4,*)$ doesn't even make sense. This has to be a typo.

Last edited: Aug 16, 2012
4. Aug 16, 2012

### kljoki

they say to try with multiplication to see what is gonna happen

5. Aug 16, 2012

### kljoki

thanks for the answer and explanation :)

6. Aug 16, 2012