# If a finite group $G$ acts transitively on a set of order $p^m$, then so does any $p$-Sylow subgroup

Here is an algebra qualifying exam problem:

Let $$G$$ be a finite group acting transitively on a set $$X$$ with cardinality $$p^m$$ for some prime $$p$$ and nonnegative integer m. Show that any $$p$$-Sylow subgroup of $$G$$ acts transitively on $$X$$.

There are some easy facts from the orbit-stabilizer theorem (a $$p$$-Sylow subgroup has order at least $$p^m$$, for example), but I don't see an obvious way to connect this to the $$P$$-orbits.

How can I solve this?

From the orbit-stabilizer theorem we have $$|G| = p^m |S_G(x)|$$, where $$S_G(x)$$ is the stabilizer in $$G$$ of any $$x \in X$$. Write $$|S_G(x)| = |G| / p^m = p^k \cdot n$$, where $$n$$ is coprime with $$p$$. Then any $$p$$-Sylow subgroup has order $$p^{m + k}$$.

Let $$P$$ be a $$p$$-Sylow subgroup of $$G$$. Again, by orbit-stabilizer, the $$P$$-orbit of an element $$x \in X$$ has cardinality $$\frac{|P|}{|S_P(x)|} = \frac{p^{m + k}}{|S_P(x)|}.$$ Since $$S_P(x)$$ is a subgroup of $$S_G(x)$$ and $$P$$, we have $$|S_P(x)| \leq p^k$$, thus the orbit has size at least $$\frac{p^{m + k}}{p^k} = p^m = |X|,$$ so $$P$$ acts transitively on $$X$$.

(This is basically a nice idea tucked away in an old comment by Derek Holt. I thought it deserved to be its own answer somewhere.)

• How do I know that $p$ is also a prime divisor of $|S_G(x)|$?
– user750041
Commented Jun 23, 2020 at 21:58
• You don't know that, and it is not assumed in the proof above. Commented Jun 23, 2020 at 22:00
• Isn't that assumed in "Write $|S_G(x)| = |G| / p^m = p^k \cdot n$"?
– user750041
Commented Jun 23, 2020 at 22:06
• @user750041 It's possible that $k = 0$, so that $p$ does not divide $|S_G(x)|$. Commented Jun 23, 2020 at 22:49
• @RobertD-B, right, nowhere is assumed $k>0$. Thanks.
– user750041
Commented Jun 23, 2020 at 23:25