When do Sylow subgroups of the same order have trivial intersection?

I'm curious because I recently read a proof where it was computed that there are $8$ Sylow 7-subgroups, and hence $8\cdot 6=48$ elements of order 7. This seems to assume that each of the Sylow subgroups has trivial intersection.

Why is this the case? Is it true even when the Sylow groups have order a prime power, not necessarily just prime order?


Two subgroups of order $7$ are either equal or intersect trivially. Indeed, their intersection is a subgroup of each of them, and they have exactly two subgroups.

In general, a group $G$ can very well have many $p$-Sylow subgroups intersecting non-trivially. For example, let $H$ be any group which does not have a normal $p$-Sylow, and consider the group $G=\mathbb Z_p\times H$.

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  • $\begingroup$ So this only works when the Sylow subgroups have prime order. Thanks. $\endgroup$ – Daniel Sep 22 '11 at 21:49
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    $\begingroup$ @Daniel: I did not say that. There are groups whose $p$-Sylows have order a proper power of $p$ and which intersect trivially in pairs. $\endgroup$ – Mariano Suárez-Álvarez Sep 22 '11 at 21:50
  • $\begingroup$ Ok, I get it now. $\endgroup$ – Daniel Sep 22 '11 at 21:55
  • $\begingroup$ To say even more: Sometimes you have three Sylow subgroups $P$, $Q$, and $R$, and $P\cap Q$ is non-trivial yet $P\cap R$ is trivial. $\endgroup$ – user641 Sep 22 '11 at 23:08
  • $\begingroup$ @MarianoSuárez-Alvarez didn't you mean $G=\mathbb Z_p\times H$? $\endgroup$ – tacos_tacos_tacos Dec 15 '12 at 23:20

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