If I have a group $G$ and a set $X$ then I can define an action $\alpha:G \times X \rightarrow X$ like $(g, x)=gx$
I can also define it's corresponding morphism $\alpha':G \rightarrow S_X$. Does this have to be suryective?
In the exercise, I have that $p$ is the lowest prime that divides $|G|$, I have an $H$ such that $|G:H|=p$ and $X=G/H$.
I tried thinking an example with something not so abstract, like $\mathbb{Z}_{15}$. So, $p=3$, $H \cong \mathbb{Z_5}$, and $X=\{\bar{0}, \bar{1}, \bar{2}\}$. $S_X = \{id, (\bar{0} \text{ } \bar{1}), (\bar{0} \text{ } \bar{2}), (\bar{1} \text{ } \bar{2}), (\bar{0} \text{ } \bar{1}\text{ } \bar{2}), (\bar{0} \text{ } \bar{2}\text{ } \bar{1})\}$.
If I have a homomorphism from $\mathbb{Z}_{15}$ to $S_X$, then, trivially, $\alpha'(0) = id$. Since $1 \in \mathbb{Z}_{15}$ has order 15, then it has to go to an element from $x \in S_X$ such that $ord(x) \text{ }| \text{ } 15$, so $ord(x) = 3$ or $ord(x) = 5$. But $S_X$ has no elements of order $5$, then it has to go to either $(\bar{0} \text{ } \bar{1}\text{ } \bar{2}) \text{ or } (\bar{0} \text{ } \bar{2}\text{ } \bar{1})$. I can apply the same reasoning to every element, so $Im(\alpha')= \{id,(\bar{0} \text{ } \bar{1}\text{ } \bar{2}), (\bar{0} \text{ } \bar{2}\text{ } \bar{1})\} $ so it wouldn't be suryective.
Is this right?
No, the morphism has not to be surjective. Consider for example $\alpha: \Bbb Z_2\times \Bbb R\to\Bbb R$ defined as $\alpha(0,x)=x$ and $\alpha(1,x)=-x$.
Then the morphism $\alpha':\Bbb Z_2\to S_{\Bbb R}$ is not surjective.