Cyclic Sylow subgroup for least prime divisor has normal complement

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 fact about::p-nilpotent group.

Related facts

 * Burnside's normal p-complement theorem
 * Cyclic normal Sylow subgroup for least prime divisor is central
 * Normal of least prime order implies central

Applications

 * Order is twice an odd number implies subgroup of index two

Facts used

 * 1) uses::Sylow satisfies intermediate subgroup condition
 * 2) uses::Cyclic normal Sylow subgroup for least prime divisor is central
 * 3) uses::Burnside's normal p-complement theorem

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).

Expository references

 * : Page 316 (relative page 2).