Intuitive meaning of transitive action

Solution 1:

To answer the title of your question, you can think of a transitive group action on a set $A$ as an action for which you can go from any point to another by the action of the group.

In general, this doesn't happen. If $G$ is a group acting on a set $A$, and $a \in A$, let $Ga$ denote all the points of $a$ you can get to by applying elements of $G$. Literally, $Ga = \{ g.a : g \in G \}$. Then in general, you will have elements $a, a', a''$ etc. (possibly uncountably many) such that $A$ is the disjoint union of the sets $Ga, Ga', Ga''$ etc. These sets are called the orbits of $A$ under the action of $G$. If there is just one set in this disjoint union, that's transitive.

If $a \in A$, then you can always restrict the action of $G$ to the set $Ga$. Then $G$ does act transitively on $Ga$.

Example:

1 . $G = \textrm{GL}_2(\mathbb R)$, the group of $2$ by $2$ real matrices with nonzero determinant. $G$ acts on $\mathbb{R}^2 - \{(0,0)\}$ by $\begin{pmatrix} a & b \\ c & d \end{pmatrix} . (x,y) = (ax + by, cx + dy)$. Since you can get from any given nonzero vector to another by applying some invertible matrix, the action here is transitive.

2 . $G = \textrm{SO}_2(\mathbb R)$, the group of $2$ by $2$ real matrices $g$ such that the determinant of $g$ is $1$ and such that $gg^t = I$, where $g^t$ is the transpose of $g$ and $I$ is the identity matrix. $G$ acts on $\mathbb{R}^2 - \{(0,0) \}$ by the same formula. You can show every $g \in G$ takes the form $$g = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$$

for some $\theta \in \mathbb{R}$, and if $v \in \mathbb{R}^2$ is a nonzero, then $g.v$ is the vector obtained by rotating $v$ counterclockwise by $\theta$ radians. In this case, you definitely cannot get from one given element to another by applying something in $G$, since you can only use $G$ to rotate points around. So the action is not transitive. It follows that the orbits of $\mathbb{R}^2 - \{(0,0\}$ under the action of $G$ are the circles with center $(0,0)$, and $\mathbb{R}^2 - \{(0,0)\}$ is the disjoint union of the orbits $$\mathbb{R}^2 - \{(0,0)\} = \bigcup\limits_{r \in (0,\infty)} \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 = r^2 \}$$

Solution 2:

Mathematicians like to copy names from different topics although their relation is ever so subtle. You should be aware of that because it can get confusing, especially when there is a generalization of a concept.