Sylow theorems

Three theorems on maximal $p$-subgroups in a finite group, proved by L. Sylow [1] and playing a major role in the theory of finite groups. Sometimes the union of all three theorems is called Sylow's theorem.

Let $G$ be a finite group of order $p^{m}s$, where $p$ is a prime number not dividing $s$. Then the following theorems hold.

Sylow's first theorem: $G$ contains subgroups of order $p^i$ for all $i=1,\dots ,m$; moreover, each subgroup of order $p^{i-1}$ is a normal subgroup in at least one subgroup of order $p^i$. This theorem implies, in particular, the following important results: there is in $G$ a Sylow subgroup of order $p^m$; any $p$-subgroup of $G$ is contained in some Sylow $p$-subgroup of order $p^m$; the index of a Sylow $p$-subgroup is not divisible by $p$; if $G=P$ is a group of order $p^m$, then any of its proper subgroups is contained in some maximal subgroup of order $p^{m-1}$ and all maximal subgroups of $P$ are normal.

Sylow's second theorem: All Sylow $p$-subgroups of a finite group are conjugate.

For infinite groups the analogous result is, in general, false.

Sylow's third theorem: The number of Sylow $p$-subgroups of a finite group divides the order of the group and is congruent to one modulo $p$.

For arbitrary sets $\pi$ of prime numbers, analogous theorems have been obtained only for finite solvable groups (see Hall subgroup). For non-solvable groups the situation is different. For example, in the alternating group $A_5$ of degree 5, for $\pi =\{2,3\}$ there is a Sylow $\pi$-subgroup $S$ of order 6 whose index is divisible by a number from $\pi$. In addition, in $A_5$ there is a Sylow $\pi$-subgroup isomorphic to $A_4$ and not conjugate with $S$. The number of Sylow $\pi$-subgroups in $A_5$ does not divide the order of $A_5$.

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