Let $C$ be an abelian additive group and write e for a generator of $C$. The elements of $C$ are then $0,e,2e,3e,\dots,(n-1)e$. If $C$ is finite, prove that the element $ke$ is another generator of $C$ if and only if $k$ and $n$ are relatively prime.
Generators of a finite additive cyclic group
2
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abstract-algebra
finite-groups
cyclic-groups
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0I do not believe so. – 2012-11-12
2 Answers
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Here is one direction: Let $\gcd(k,n) = 1 $ then there exist $i, j \in \mathbb Z$ such that $1 = ik + jn$. If $me$ is an arbitrary element in $C$, $m \in \mathbb Z$, then $me = m(ik + in) e = mik e + min e = mike = (mi)ke$.
The statement is a actually a direct consequence of the following theorem:
Let $a$ be an element of order $n$ and let $k$ be a positive integer. Then $| a^k| = n / \gcd(n,k)$.
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0See for example [Gallian](http://books.google.ch/books/about/Contemporary_Abstract_Algebra.html?id=CnH3mlOKpsMC&redir_esc=y), 7th Edition, page 75, Theorem 4.2. – 2012-11-12
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Suppose gcd$(n, k) = 1$. Supose $mke = 0$ for an integer $m$. Then $n|mk$. Since gcd$(n, k) = 1$, $n|m$. Hence the order of $ke$ is $n$.
Conversely suppose $d =$ gcd$(n, k) \ne 1$. Let $k' = \frac{k}{d}, n' = \frac{n}{d}$. Then $n'ke = n'dk'e = nk'e = 0$. Since $n' < n, ke$ is not a generator.
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0Can you elaborate why from $mke = 0 $, $n | mk$ follows? – 2018-12-24