Generating Z 10

[corbell777]

I am studying on my own, so have no teacher, and I got stumped! Please help.

The book asked me to find the generators for Z 10, which I did. They are {1}, {3}, {7}, and (9}. Then the very next problem asked me to generate Z 10 with 2 and 5. I thought maybe they meant start with 2 and add by 5's, but that gives 7, 2, 7, 2, 7, 2, ad infinitum. So then I thought it might be start with 5 and keep adding 2's, but that's just generating by {2} only starting with 5 instead of 0. In that case, I could start with ANY number and generate it by 2's. Just exactly what are they asking me to do?

Thank you for any clarification you can give me.

 

[daon2]

I am studying on my own, so have no teacher, and I got stumped! Please help.

The book asked me to find the generators for Z 10, which I did. They are {1}, {3}, {7}, and (9}. Then the very next problem asked me to generate Z 10 with 2 and 5. I thought maybe they meant start with 2 and add by 5's, but that gives 7, 2, 7, 2, 7, 2, ad infinitum. So then I thought it might be start with 5 and keep adding 2's, but that's just generating by {2} only starting with 5 instead of 0. In that case, I could start with ANY number and generate it by 2's. Just exactly what are they asking me to do?

Thank you for any clarification you can give me.

Since GCD(2,5)=1, every integer is a linear combination of 2's and 5's. Hence they generate Z_n for any n. First note that 2(-2) + 5(1)=1. In Z_10, -2 = 8. Hence 2*(8)+5*(1) = 1 (mod 10).

You can get any value N (mod 10) by multiplying this relation by N (mod 10).

 

[girodd]

probably they want you to recognize that Z10 is the direct sum of Z2 and Z5. You would start with the multiples of 5

= H = {0,5}

and the multiples of 2

= K ={0,2,4,6,8},

each of which is a cyclic subgroup of Z10, and note that every element of Z10 can be expressed in exactly one way as a sum of an element of H and an element of K