# $f(x) = x^r$ equaling to $r \cdot x$ when $f$ is cyclic group $G \rightarrow G$

Suppose that there is a cyclic group $G$ of order $n$ with a generator $g$. Also suppose that $r$ is a fixed integer. Then define a homomorphism function $f: G \rightarrow G$, $f(x) = x^r$.

How do we get that $f(x) = r \cdot x$?

Quote:

Theorem: Let $G$ be a cyclic group of order $n$ with generator $g$. Fix an integer $r$, and define $f : G \rightarrow G$ by $f(x) = x^r$. This map $f$ is a group homomorphism of $G$ to itself. If $gcd(r, n) = 1$, then $f$ is an isomorphism, and in that case every $y ∈ G$ has a unique $r$th root. More generally, order of kernel of $f = gcd(r, n)$, order of image of $f = n/gcd(r, n)$. If an element $y$ has an $r$th root, then it has exactly $gcd(r, n)$ of them. There are exactly $n/gcd(r, n)$ $r$th powers in $G$. Proof: Since $G$ is abelian the map $f$ is a homomorphism. Use the fact that $G$ is isomorphic to $\mathbb{Z}/n$. Converting to the additive notation for $\mathbb{Z}/n$-with-addition, f is $f(x) = r · x$

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I must be missing something. What do you mean by $r \cdot x$? Is $r$ an element of $G$? – JohnJamesSmith Aug 18 '12 at 4:17
@JohnJamesSmith no. $r$ is just a fixed integer. – Lucy Zeo Aug 18 '12 at 4:21
The added quote clarifies everything :) When the group operation is addition, $f(x) = rx$ by definition. As Marc notes, the notation is somewhat distasteful. – JohnJamesSmith Aug 18 '12 at 4:32
I am slightly getting confused. So, $\cdot$ is additive operator? (which does not make sense to me) Or is it something else? – Lucy Zeo Aug 18 '12 at 4:36
No, the $\cdot$ in $r\cdot x$ is plain old multiplication of integers. This multiplication is not the same as the group operation, which in this case happens to be integer addition. – JohnJamesSmith Aug 18 '12 at 4:38

The notation $r\cdot x$ is just the same as $x^r$, but using additive notation for $G$, which is appropriate if (for instance) $G=\mathbf Z/n\mathbf Z$. You should avoid mixing additive and multiplicative notation for the same group, but it happens that some general fact written multiplicatively applies to a particular group written additively, and you'll have to cope with translating notation in that case.