Let $G=Z_n$. How can I show that the number of generators $G$ is $\varphi(n)$ whereas $\varphi$ is Euler's totient function?
I know That I need to show here a start of a solution but all my edges had diverged.
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But what's your $n$? and I think Lagrange's theorem will do. – linear_combinatori_probabi Oct 30 '18 at 15:53
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Your question makes no sense the way it is written. Without you telling me where $x$ "lives", the only reasonable conclusion is that $G = \langle x \rangle$ is the free group on one generator. – Sam Streeter Oct 30 '18 at 15:53
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@SamStreeter What do you mean by "where x lives"? x is some organ in the group $G$. – J. Doe Oct 30 '18 at 15:54
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1@J.Doe My comment was made before the question was edited. Originally, it just read $G = \langle x \rangle$. – Sam Streeter Oct 30 '18 at 16:18
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@J.Doe Also, I've never heard "organ" used in place of "element". Where can I find this terminology being used, out of interest? – Sam Streeter Oct 30 '18 at 16:19
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Let $1\le d \le n$, $d$ integer. Suppose $\gcd(d,n)=k>1$, so that you can write $d=kd_0$, $n=kn_0$, with $\gcd(d_0,n_0)=1$. Then ${(x^d)}^{n_0}=(x^{n})^{d_0}=1^{d_0}=1$.
If $k>1$, then $n_0<n$, so $x^d$ is not a generator in this case. This shows that the number of generators is at most $\phi(n)$.
Suppose now that $k=1$. Then there exists a unique integer $1\le r \le n$ such that $d\cdot r \equiv 1 \mod n$. It follows that for every $1\le m \le n$ you have $$ (x^d)^{mr}= (x^{dr})^m =x^m.$$ Therefore in this case $x^d$ is a generator and you conclude that the number of generators is exaclty $\phi(n)$.
