Size of the deck transformation group

Question

If $p\colon Y\to X$ is a $k$-fold covering map, and $Y$ is path-connected, what is the size of Deck($p$), the deck transformation group?

I was attempting to prove that the answer is $\leq k$, but thought of an example where I think it is $k!$, so my question is - where have I gone wrong?

image borrowed from this question.

In particular it is this example from Hatcher's Algebraic Topology, the three-fold cover of the figure 8 graph pictured above. Could we not use any different combination of $a$ or $b$ to permute the vertices (labeled $v_1,v_2,v_3$ from left-to-right) into any order? i.e. can the deck group be isomorphic to $S_3$ with elements:

1. $e$
2. $a$ - fixes $v_1$, switches $v_2$ and $v_3$
3. $b$ - fixes $v_3$, switches $v_1$ and $v_2$
4. $bab$ - fixes $v_2$, switches $v_1$ and $v_3$
5. $ab$ - $v_1\mapsto v_2$, $v_2\mapsto v_3$, $v_3\mapsto v_1$
6. $ba$ - $v_1\mapsto v_3$, $v_2\mapsto v_1$, $v_3\mapsto v_2$

I realize I am wrong, but would like to know why to help my understanding of deck groups, thanks

Answer 1

We know that any $f\in\text{Deck}(p)$ fixes some $x\in Y$ if and only if $f=id_Y$. Suppose that the size of $\text{Deck}(p)$ is greater than $k$. Then by the pigeonhole principle, there are $\tau,\sigma\in\text{Deck}(p)\setminus id_Y$ such that $\tau({V_1})=\sigma({V_1})=V_i$, without loss of generality, where $V_1,V_i,\,i\leq k$ are disjoint subsets of $Y$ that are mapped homeomorphically onto $X$ by $p$, as in the definition of covering spaces.

We have that for all $x\in V_1$, $p\circ\tau(x)=p(x)$, and $p\circ\sigma(x)=p(x)$, and given that $p|_{V_i}$ is a homeomorphism, we must have $\tau(x)=\sigma(x)$, a contradiction.

Answer 2

The Deck transformation group acts freely on $Y$ and preserves the fibers so its cardinal is inferior to the cardinal of the fiber