# Cyclic Sylow subgroup for least prime divisor has normal complement

This article gives the statement, and possibly proof, of a normal p-complement theorem: necessary and/or sufficient conditions for the existence of a Normal p-complement (?). In other words, it gives necessary and/or sufficient conditions for a given finite group to be a P-nilpotent group (?) for some prime number $p$.
View other normal p-complement theorems
This result relates to the least prime divisor of the order of a group. View more such results

## History

This result is generally attributed to Burnside. It appeared in a paper by him published in 1895.

## Statement

Suppose $G$ is a finite group and $p$ is the least prime divisor of the order of $G$. If a $p$-Sylow subgroup of $G$ is cyclic, then it has a normal complement: in particular, it is a retract. In symbols, if $P$ is a cyclic $p$-Sylow subgroup of $G$, then there exists a normal subgroup $N$ of $G$ such that $NP = G$ and $N \cap P$ is trivial. Another way of putting this is that $G$ is a P-nilpotent group (?).

## Proof

Given: A finite group $G$. $p$ is the least prime divisor of the order of $G$, and $P$ is a $p$-Sylow subgroup of $G$.

To prove: $P$ has a normal complement in $G$.

Proof:

1. $P \le Z(N_G(P))$: Consider $P$ as a subgroup of its normalizer $N_G(P)$. Note that $p$ is still the least prime divisor of the order of $N_G(P)$, and by fact (1), $P$ is $p$-Sylow in $N_G(P)$. Thus, by fact (2), $P \le Z(N_G(P))$.
2. $P$ has a normal complement: This follows from the conclusion of the previous step and fact (3).