Order of an element in a group

[Dick]

I am sorry to rehash this again, but I am having trouble with one argument I made:
I don't see why this implies that $a$ and $k$ have to have the same exponent. Isn't it possible that $k > a$, yet $g^k$ and #g^a## get mapped to the same element?

Correct. I didn't read closely enough. $g^k=g^a$ doesn't mean k and a are the same. Just the powers of g are the same. Omit that statement and the rest is fine.

 

[Bashyboy]

I am not sure how I can conclude $\langle g \rangle$ and $G_r$ are the same set, then. Haven't I only demonstrated that $G_r$ is a subset?

 

[Dick]

I am not sure how I can conclude $\langle g \rangle$ and $G_r$ are the same set, then. Haven't I only demonstrated that $G_r$ is a subset?

Haven't you shown that for any a, $g^a=g^k$ for some k in {0,1,...,r-1}? Isn't that enough?

 

[Bashyboy]

Oh, and so this forces them to be equal, and it forces the exponents of $\langle g \rangle$, without any repetition, to be {0,1,2,3...,r-1}. Is that right?

 

[Dick]

Oh, and so this forces them to be equal, and it forces the exponents of $\langle g \rangle$, without any repetition, to be {0,1,2,3...,r-1}. Is that right?

What do you mean 'without repetition'? $g^a$ is in $G_r$ for any a. So $\langle g \rangle$ is contained in $G_r$. Isn't that clear?