Index of cheque generator rex11/30/2022 ![]() And every element of the group has the form $g^m$ for some $m$. ![]() So for $g^m$ to be a generator - or equivalently, for $g$ to have order $n$ - it is both necessary and sufficient that $m$ be relatively prime to $n$. It is a fact, which I encourage you to prove if you have not encountered it, that for an integer $m$ between $1$ and $n$ the order of $g^m$ is $n/(m,n)$, where $(m,n)$ is the greatest common denominator of $m$ and $n$. Why is my claim correct? Suppose $g$ is a generator for the group, so that $g$ has order $n$. If your cyclic group has order $n$, I claim that there will be one generator for every number between $1$ and $n-1$ (inclusive) that is relatively prime to $n$: in other words, there are $\varphi(n)$ generators, where $\varphi$ is Euler's totient function. But much of the time when you work with a cyclic group you will also naturally know of a generator. ![]() For example, I believe there is no fast algorithm to find a generator for the multiplicative group $(\mathbb Z/p^k\mathbb Z)^\times$ when $p$ is a large prime. In the general case, finding the generator of a cyclic group is difficult. ![]()
0 Comments
Leave a Reply.AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |