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Let $G$ be a group of order $12$. In an example from Dummit and Foote that classifies groups of order $12$, it says that each Sylow $3$-subgroup intersects in the identity, and if $n_3=4$, then $G$ contains $2\cdot 4=8$ elements of order $3$.

How are we certain that Sylow $3$-subgroups have only the identity as a common element?

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    $\begingroup$ In $G$, each Sylow 3-subgroup is a copy of $\mathbb{Z}_3$. Do you see why these can only intersect trivially? What would happen if they shared just one non-identity element? $\endgroup$ – Ashwin Trisal Oct 30 '17 at 0:24
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    $\begingroup$ They're all $\mathbb{Z}_3$s. If they intersect nontrivially they're equal. $\endgroup$ – Randall Oct 30 '17 at 0:25
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If you are given a group $G$ with two distinct subgroups, say $P,P'\leq G$ of (the same) prime cardinality $p=|P|=|P'|$ you have that $\left|P\cap P'\right|\mid\left|P\right|=p$ so being $P\cap P'\neq P$ (as it would imply $P=P'$) we must have $\left|P\cap P'\right|=1$ so $P\cap P'=\left\{1\right\}$

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