Homework Help: 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

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