0
$\begingroup$

Let's say we have $5$ Sylow $p$-subgroups in $G$ for some $p$. This induces a group homomorphism $f: G\to S_{5}$ if we let $G$ act on the set $Syl_p$ by conjugation.

Question: I'm not 100% sure if I understand why it follows that $|f(G)|\geq 5$ from the fact that the group action is transitive.

I would say that we have at least permutations $\sigma$ with $\sigma (i)=j$ for each $i,j\in\{1,\dots,5\}$. So for example, if we fix $i=1$ we get 5 permutations for this that cannot be the same since they are bijections. Is that correct? Maybe someone else has more insight and would care sharing his point of view?

$\endgroup$
  • 1
    $\begingroup$ So $p=2{{{}}}$. $\endgroup$ – Lord Shark the Unknown Dec 13 '17 at 21:09
  • $\begingroup$ What do you mean? $\endgroup$ – Buh Dec 13 '17 at 21:12
  • $\begingroup$ $p=2$ is the only prime that satisfies $5 \equiv 1 \bmod p$. $\endgroup$ – Derek Holt Dec 13 '17 at 23:04
  • $\begingroup$ Yeah but that has nothing to do with my question. $\endgroup$ – Buh Dec 14 '17 at 8:42
1
$\begingroup$

Just use the class formula: there results from this formula that, for any group $G$ operating on a set $X$, the cardinal of an orbit is a divisor of $|G|$.

$\endgroup$
  • 1
    $\begingroup$ Okay, so all Sylow-$p$-subgroups are in the same Orbit with length $5$. But how does this mean that the image of $G$ has at least 5 elements? Maybe it's too late and I'm too tired. $\endgroup$ – Buh Dec 13 '17 at 20:41
  • 1
    $\begingroup$ If $5$ is a divisor of $|f(G)|$, it implies $|f(G)|\ge 5$, don't you think? $\endgroup$ – Bernard Dec 13 '17 at 20:48
  • $\begingroup$ Didn't we just say $5$ is a divisor of $|G|$? $\endgroup$ – Buh Dec 13 '17 at 21:11
  • $\begingroup$ In my answer, it was a general notation (for any group $G$...). It is applied to your $f(G)$ as a subgroup of $S_5$. $\endgroup$ – Bernard Dec 13 '17 at 21:15
  • $\begingroup$ How so? G acts on the set of Sylow subgroups, not $f(G)$. $\endgroup$ – Buh Dec 13 '17 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.