# Transitivity of group actions - examples

Suppose that the group $$G$$ acts on a set $$S$$ and $$H\le G$$. If the action of $$H$$ on $$S$$ is transitive, then so is the action of $$G$$ on $$S$$. I was wondering whether the converse is true i.e.

"If the action of $$G$$ on $$S$$ is transitive, can we conclude the same for the action of $$H$$ on $$S$$?"

My attempt:

Let $$G=D_8$$ and $$S=\{\text{vertices of a square}\}$$. The action of $$G$$ on $$S$$ is transitive, since each vertex can be mapped every other one, using one of the 8 isometries in $$G$$. Let $$H=C_2\cong\langle \sigma\rangle$$, generated by an orthogonal reflection. Each vertex is mapped to itself or to one other vertex (depending on what reflection is chosen). This proves that the transitivity of $$G$$ on $$S$$ does not imply the transitivity of $$H$$ on $$S$$.

Is this proof acceptable?

Are there other examples (i.e. no dihedral groups)?

Thanks.

Let $$G$$ be any group with $$|G| > 1$$ and $$X$$ be any set with $$|X| > 1$$. Suppose $$G$$ acts transitively on $$X$$.
Consider $$H = \langle e\rangle,$$ the trivial subsgroup.
Given any $$x \in X$$, we have that $$e\cdot x = x$$ and thus, the restricted action is not transitive as $$|X| > 1$$.
In fact, as long as you take a subgroup $$H \le G$$, such that $$|H| < |X|,$$ you can directly conclude that the restricted action is not transitive.
• To be complete at this level of detail, you have to say that there indeed exists a group that acts transitively on a set of cardinal $>1$...! – YCor Jan 10 '20 at 14:43