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I'm having a hard time grasping the concept of finding the number of Sylow p-subgroups. Consider a group $G$ of order $3 \cdot 5 \cdot 7 = 105$. Now, let's find the number of Sylow 3-subgroups for this group: we have

$5 \equiv 2 \pmod 3\\7 \equiv 1 \pmod 3\\35 \equiv 2 \pmod 3$

So $1$ and $7$ are my options, by Sylow's third theorem. How do I determine which, in a none-bruteforce manner? (That is, I consider "we found $2$ so there must be $7$" a bad method. If you disagree strongly about this, feel free to present your argument and convince me, as long as you present a convenient method for the case where it is $1$).

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You do need to bruteforce insofar as the answer depends on details of the group. We could have $G\cong (Z_7\rtimes Z_3)\times Z_5$ where there are seven and we could have $G\cong Z_{105}$ where there is just one.

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  • $\begingroup$ Just to be clear: given that we only know the order of the group, there is no way to determine whether its $1$ or $7$ in our scenario? $\endgroup$ – Andrew Thompson Apr 25 '14 at 17:31
  • $\begingroup$ Yes, Exactly. I didnÄt check though if the number of $5$-groups or the number of $7$-groups is fully determined. Or if there are still more groups ... $\endgroup$ – Hagen von Eitzen Apr 25 '14 at 17:35
  • $\begingroup$ Well, they may only be fully determined if there is only one if we only know the order of the group, if I understand you correctly. $\endgroup$ – Andrew Thompson Apr 25 '14 at 17:42
  • $\begingroup$ For a group of order $105$, the number of $5$-Sylows is exactly $1$, same for $7$-Sylows. $\endgroup$ – Mikko Korhonen Apr 25 '14 at 17:47
  • $\begingroup$ @MikkoKorhonen, why isn't $21$ an alternative for $5$-Sylow? $7 \cdot 3 \equiv 1 \pmod 5$. $\endgroup$ – Andrew Thompson Apr 25 '14 at 19:45

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