I would like to solve the Problem 2.19 from A Course in Modern Mathematical Physics by Szekeres. The problem is part of the paragraph 2.6 Group action. The formulation is:
Problem 2.19 If $H$ is any subgroup of a group $G$ define the action
of $G$ on the set of left cosets $G/H$ by $g$: $g'H\mapsto gg'H$.(a) Show that this is always a transitive action of $H$ on $G$.
In the formulation in (a) this refer to the action mentioned before:
$G\to \text{Transf}(G/H),g\mapsto(g'H\mapsto gg'H).$
But the action mentioned in (a) is:
$H\to \text{Transf}(G).$
That is the two actions are not the same.
What am I missing? How should I understand the problem?
Note that I am not asking about the proof that the action is transitive, I am interested in the problem formulation.
Best Answer
The action is of $G$ on $G/H$, where $g(g'H) = (gg')H$. This a trasitive group action (one orbit) since for any $gH,g'H \in G/H$, let $h = g'g^{-1}$, then $h(gH) = g'H$. Equivalently, for any $gH \in G/H$, $g(H) = gH$.