A silly confusion with permutations? I'm reading Beardon's Algebra and Geometry, I'm a little confused about this exercise: 



If $n$ is not divisible by $3$, then $\rho$ fixes some $k$. Suppose $n=5$, then we have the set $\{1,2,3,4,5\}$. As $\rho$ is a $3-$ cycle, we write it as - for example:
$$\begin{pmatrix}
{1}&{2}&{3}\\ 
{2}&{3}&{1}
\end{pmatrix}$$
Or
$$\begin{pmatrix}
{1}&{2}&{3}&{4}&{5}\\ 
{2}&{3}&{1}&{4}&{5}
\end{pmatrix}$$
Which amounts to the same thing. Now suppose we apply it to $12345$, then we will have $23145$ then it obviously will have $4,5$ fixed. But suppose $n=6$, we apply the same permutation to $123456$ and obtain $231456$ and it also happens that some integers are fixed, also if the permutation is a $3-$cycle, it obviously can't reach all the integers in the permutation. I may have understood something wrong. 
 A: Write $\rho$ as a product of disjoint cycles: $\rho=\sigma_1\sigma_2\dots\sigma_k$. Suppose that $\rho^s=I$. Then, by the properties of disjoint cycles,
$$
\sigma_1^s \sigma_2^s \dots \sigma_k^s = I
$$
If $\sigma_i^s\ne I$, then there is $p\in\{1,\dots,n\}$ such that $q=\sigma_i^s(p)\ne p$. On the other hand, $q$ is not moved by any $\sigma_j^s$, because of disjointness. This is a contradiction to the product being the identity.
Therefore $\sigma_i^s=I$, for $i=1,\dots,k$. For $s=3$ this implies that the length of $\sigma_i$ is either $1$ or $3$.
Without loss of generality, we can assume $\sigma_i$ has length $3$ for $1\le i\le h$ and length $1$ for $h+1\le i\le k$. Then
$$
n=3h+(k-h)
$$
If $n$ is not divisible by $3$, we conclude that $k-h\ne0$, which means that $\sigma_k$ has length $1$ and at least one element is fixed by $\rho$: indeed, $k-h$ is the number of length $1$ cycles, that is, of fixed elements.
Nothing can be said about fixed elements in case $n$ is divisible by $3$. There can be fixed elements or not. For instance, if $n=6$, we can consider
$$
\rho_1=(123)(4)(5)(6)
\qquad\text{and}\qquad
\rho_2=(123)(456)
$$
The former has fixed elements, the latter hasn't. What one can say is that the number of fixed elements is also divisible by $3$, because, with the same notation as before, $k-h=n-3h$.
