# A transitive subgroup H of $S_n$

I have to prove the following:

A transitive subgroup $H$ of $S_n$ is primitive if and only if for all $x\in X$, we have that $H_x:=\{ \sigma \in H : \sigma(x)=x \}$ is a maximal subgroup of H (i.e., there does not exist a group $H'$ with $H_x \subset H' \subset H$,$H' \neq H_x$ and $H'\neq H$).

Can someone please explain what it means for a subgroup to be transitive and primitive? I don't understand the definition given by my teacher.

The two concepts are related. Presumably you know the definition of the left regular action of a group on itself: $g\cdot h = gh$. Given that, a subgroup $H$ is transitive if $H$ acts transitively on the entire group $G$. This means that for any pair of elements $g_1, g_2\in G$, there is some element $h\in H$ such that $h\cdot g_1 = g_2$.
Suppose that $G = G_1\cup\cdots \cup G_n$ is a partition of $G$ into disjoint subsets. It is possible that the left regular action also acts on these subsets; that is, that for every $h\in H$ and each $G_i$, we have $h\cdot G_i = G_j$ for some $j$. Clearly this is true if the $G_i$ are the collection of all elements in $G$, and it is also true if $G_1=G$ is the partition into a single subset. If this condition does not hold for any other partition, the subgroup is said to be primitive. So: a subgroup is primitive if, for any partition of $G$ other than the trivial partitions, it does not act on that partition since it mixes up the group elements from different elements of the partition.