$\mathbb{Z}_5$ as a subgroup of $S_5$ and the induced action on a vector space.

Question

I'm just trying to understand the following. Consider the cyclic group $(\mathbb{Z}_5,+)$ and the vector space $\mathbb{R}^5$.

First Question:

Can we say that $\mathbb{Z}_5 \le S_5$ (it is a subgroup of $S_5$ generated by the cycle $(01234))$ ?

And if yes, how to see that?

Second Question:

Also let's define an action by

\begin{align} \mathbb{Z}_5 \times \mathbb{R}^5 &\rightarrow \mathbb{R}^5\\ (g,\textbf{v}) &\mapsto g \cdot \textbf{v}:= (v_{g(1)}, v_{g(2)}, v_{g(3)},v_{g(4)}, v_{g(5)}) \end{align}

That is, we permute the coordinates of the vector $\textbf{v} \in \mathbb{R}^5$. How many of those permutations we can find and why?

I think that the answer is $5$ because but I would like to be sure of why is that. To define the above action, should I decide the dimension of the vector space accordingly to the number of elements in the group?

Many thanks,

James

Answer 1

The action is not well-defined. For this, you need the setting

$g\cdot (v_1,\ldots,v_5) := (v_{g^{-1}(1)},\ldots,v_{g^{-1}(5)}).$

In this way, $(gh)\cdot v = g\cdot (h\cdot v)$ which does not hold in the above setting.