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For example, $|G|=57=19\cdot3$

By Sylow's Theorem, $G$ has no. of Sylow 3-subgroup and Sylow 19-subgroup $n_3=19,n_{19}=1$.

Define those $19$ subgroups as $S_{3,k},1\leq k\leq19$ that have trivial intersection but can we say $$C_{19}\cap S_{3,k}=\{e\}$$?

If yes, more generally, are Sylow p-subgroups of prime order always have trivial intersection?

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  • $\begingroup$ Hint: The intersection is never empty (it will always contain the identity). Also, use Lagrange's theorem. $\endgroup$ – Tobias Kildetoft Dec 16 '16 at 9:54
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    $\begingroup$ They all have trivial intersection by Lagrange's Theorem! $\endgroup$ – ZHU Dec 16 '16 at 9:56
  • $\begingroup$ Yes, precisely (when they have prime order as here. Otherwise they need not). $\endgroup$ – Tobias Kildetoft Dec 16 '16 at 9:57
  • $\begingroup$ What does "$\,S_{3,k},\,1\le k\le 19\;$ " mean ? $\endgroup$ – DonAntonio Dec 16 '16 at 9:58
  • $\begingroup$ @DonAntonio "Define those $19$ subgroups as $S_{3,k},1\leq k\leq19$" I would interpret that as $S_3$ meaning it's a Sylow $3$-subgroup, and the $_{,k}$ to be an indexing of those $19$ groups ($k$ ranging, obviously, from $1$ to $19$, inclusive). So $S_{3,1}$ is the first one and $S_{3,19}$ is the last one. $\endgroup$ – Arthur Dec 16 '16 at 10:03
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Let $A$ and $B$ be subgroups of $G$ with $\gcd(|A|,|B|)=1$. The subgroup $A \cap B$ is subgroup of $A$ so $|A \cap B|$ divides $|A|$, by Lagrange's Theorem. Likewise $|A \cap B|$ divides $|B|$. Since $\gcd(|A|,|B|)=1$, we have that $|A \cap B|=1$. Therefore $A\cap B =\{1\}$.

Now taking $A$ to be a Sylow $p$-subgroup and $B$ to be a Sylow $q$-subgroup yields the desired result because $\gcd(|A|,|B|)=\gcd(p^r,q^s)=1$

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