I stumbled upon this nice and neat proof about how a normal $p$-subgroup contained in all Sylow $p$-subgroups.
https://proofwiki.org/wiki/Normal_p-Subgroup_contained_in_All_Sylow_p-Subgroups
But in the step where second Sylow theorem is recalled, the proof states the second Sylow theorem as:
Let $P$ be a Sylow $p$-subgroup of the finite group $G$.
Let $Q$ be any $p$-subgroup of $G$.
Then $Q$ is a subset of a conjugate of $P$.
But Wiki as most literature states this as: All Sylow $p$ subgroups are conjugate to each other. See https://en.wikipedia.org/wiki/Sylow_theorems.
How this two statements of second Sylow theorem are equal?
Best Answer
Assume the alternative statement of the second theorem. If $Q$ is large enough to be a Sylow subgroup, then assuming it's contained in a conjugate of $P$ means (since they have the same size) that it must be a conjugate of $P$.
Conversely, if all Sylow $p$-subgroups are conjugate to one another and $Q$ is a $p$-subgroup (not necessarily Sylow), then $Q$ is contained in some Sylow subgroup and that Sylow subgroup is conjugate to $P$, so $Q$ is contained in a conjugate of $P$.