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?

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    $\begingroup$ 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

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 $|G|$.

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    $\begingroup$ 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
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    $\begingroup$ 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

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