Group Homomorphism mapping to the symmetric group on X elements

Question

If we have a group G acting on a Set X define the group homomorphism$$ \varphi:G\to S_x $$ where$$\varphi(g)=\sigma_g ,\sigma_g(x)=g*x=x$$

I am trying to show that if this group homomorphism is surjective then $*$ is transitive. I'd like to see if I have my ideas/understanding clear and if there are points I am missing.

If the group homomorphism is surjective that means for some $g\in G$ we have $\varphi(g)$ that maps every to every permutation in $S_x$

Since $g*x$ is a permutation on $X$ then if we have an $x\in X$, $\varphi(g)$ will send it to itself or another element in X. Since $\varphi$ is surjective there is going to be a permutation that sends a fixed $x$ to every $y\in X$. Which makes $*$ transitive.

Also do we need the group homormorphism to be surjetive for transitivity? If $S_x$ has n elements then there are n! permutations. I believe you need less than that to have this property hold true (atleast for $3\leqslant n$.)

Answer 1

What you wrote is correct, though a bit unclear. You could make things more clear as follows:

Let $x,y\in X$. There is a permutation $\sigma\in S_X$ mapping $x$ to $y$ (one could take the permutation $(x\,\,\,\, y)$). Since $\varphi$ is surjective, there is $g\in G$ such that $\varphi(g)=\sigma$. Then we have $g*x=\varphi(g)\cdot x=(x\,\,\,\,y)\cdot x=y$ as desired.

For the second question, indeed surjectivity is not needed. In fact, it is enough for the image of $\varphi$ to act transitively on $X$. For example, if $X$ has $n$ elements, then the cyclic subgroup of $S_n$ generated by the cyclic permutation $(1\,2\,\dots\,n)$ (or, in fact, by any $n$-cycle) acts transitively on $X$, and this subgroup has only $n$ elements. Another example is the subgroup $A_n$ of even permutations, which acts transitively on $X$.