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Here is the definition of Sylow p-group (source: wikipedia)

For a prime number p, a Sylow p-subgroup (sometimes p-Sylow subgroup) of a group G is a maximal p-subgroup of G, i.e. a subgroup of G that is a p-group (so that the order of every group element is a power of p) that is not a proper subgroup of any other p-subgroup of G. The set of all Sylow p-subgroups for a given prime p is sometimes written Sylp(G).

Does that mean it is possible that if I have $2$ sylow p-groups $P_1,P_2$ with $P_1 \neq P_2$ then

$|P_1| \neq |P_2|$ (if neither $P_1 \leq P_2,$ nor $P_2 \leq P_1)$ or am I misunderstanding the definition.

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The definition would a priori allow to have $|P_1|\ne|P_2|$. However, the Sylow theorems tell us that the Sylow p-subgroups are conjugates of each other and hence of same order (in fact, isomorphic).

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  • $\begingroup$ That's the problem. In the proof of that Sylow theorem in my book, they used $|P_1| = |P_2|$. So that's why I'm asking the question. $\endgroup$ – user370967 May 15 '17 at 14:39
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    $\begingroup$ @Math_QED Presumably, your book defines a Sylow $p$-subgroup as a subgroup of order $p^a$ (where $|G| = p^a n$ with $p \nmid n$). So it follows immediately that all Sylow $p$-subgroups are maximal (among $p$-subgroups). And presumably, your books goes on to prove the theorem that any $p$-subgroup of $G$ is contained inside a Sylow $p$-subgroup. From this, it follows that all maximal $p$-subgroups are Sylow $p$-subgroups. So the two definitions are equivalent. $\endgroup$ – Kenny Wong May 15 '17 at 15:03
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    $\begingroup$ Sylow's theorems only apply to finite groups. For infinite groups it is possible to have $|P_1| \ne |P_2|$. If $P$ and $Q$ are any two $p$-groups, finite or infinite, then $P$ and $Q$ are both Sylow $p$-subgroups of the free product $P*Q$. $\endgroup$ – Derek Holt May 15 '17 at 15:09
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Some authors define Sylow $p$-subgroups having order $p^k$ with $k$ maximal, see here:

Definition: If $p^k$ is the highest power of a prime $p$ dividing the order of a finite group $G$, then a subgroup of G of order $p^k$ is called a Sylow $p$-subgroup of $G$.

With this definition, $|P_1|=|P_2|$. There is also another definition as a maximal $p$-subgroup of $G$, and then we would use a Sylow Theorem to show that $P_1=gP_2g^{-1}$ for some $g$, so that $|P_1|=|P_2|$.

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  • $\begingroup$ Yes, but that is not the definition being used in the question! The advantage of the wikipedia definition is that it makes sense for infinite groups, although for infinite groups they may not exist at all, or there may be two such groups that are not conjugate. $\endgroup$ – Derek Holt May 15 '17 at 15:06

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