Order of Group Elements: Z3 x Z3 & Z2 x Z4

[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, *[/color])
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)[/color]
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)
please help :)
thanks for your time

 

[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?

 

[christoff]

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?

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.

 

[kljoki]

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?

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

 

[kljoki]

thanks for the answer and explanation :)

 

[kljoki]

ok and what about (Z3xZ5,*)
(0,0),(0,1),(0,2),(0,3),(0,4) will have order 1 or what??
how can (aⁿ(mod 3),bⁿ(mod 5)) = (1,1) when the element have (0,b)?? :D
what should i do here??

 

[christoff]

These will have order equal to the order of the "right" element. This is not a group under multiplication (see my earlier edit - I was mistaken. Zero never has a multiplicative inverse.); only addition.