# Group Homomorphism mapping to the symmetric group on X elements

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 $$3\leqslant n$$.)

## 1 Answer

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 $$X$$.