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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.

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Yes, this proof is acceptable. In general, you can consider a much simpler example.

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.

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  • $\begingroup$ 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 $>1$...! $\endgroup$ – YCor Jan 10 '20 at 14:43

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