Order is twice an odd number implies subgroup of index two

Statement
Suppose $$G$$ is a finite group and the order of $$G$$ is $$2m$$, where $$m$$ is an odd number. Then, $$G$$ has a fact about::subgroup of index two, i.e., a subgroup of order $$m$$.

Note that since index two implies normal, the subgroup of order $$m$$ is in fact a normal subgroup, and is thus the fact about::Brauer core of the whole group.

Stronger facts

 * Burnside's normal p-complement theorem is a significant generalization of this result, and states that if a $$p$$-Sylow subgroup is in the center of its normalizer, it has a normal complement.
 * Cyclic Sylow subgroup for least prime divisor has normal complement is a generalization of the stated result, and this generalization follows from Burnside's normal p-complement theorem and the fact that Cyclic normal Sylow subgroup for least prime divisor is central.
 * Frobenius' normal p-complement theorem

For a complete list of normal p-complement theorems, refer:

Category:Normal p-complement theorems

Related facts about index two and least prime index

 * Index two implies normal
 * Subgroup of least prime index is normal: If $$p$$ is the least prime dividing the order of a finite group $$G$$, any subgroup of index $$p$$ is normal.
 * Normal of least prime order implies central

Facts used

 * 1) uses::Cayley's theorem: Any group $$G$$ can be embedded inside the symmetric group $$\operatorname{Sym}(G)$$, where an element $$g \in G$$ acts by left multiplication.
 * 2) uses::Cauchy's theorem: If a prime $$p$$ divides the order of a finite group $$G$$, then $$G$$ has an element of order $$p$$.
 * 3) uses::Index satisfies transfer inequality: If $$A,B$$ are subgroups of $$C$$, then $$[A:A \cap B] \le [C:B]$$.

Elementary proof
Given: A finite group $$G$$ of order $$2m$$, where $$m$$ is an odd integer.

To prove: $$G$$ has a subgroup of index two.

Proof: By fact (1), consider the embedding of $$G$$ as a subgroup of $$K = \operatorname{Sym}(G)$$. Let $$L = \operatorname{Alt}(G)$$ be the alternating group on $$G$$. By definition $$L$$ is a subgroup of index two in $$G$$.


 * 1) $$G$$ contains an element, say $$g$$, of order two: This follows from fact (2).
 * 2) $$g$$, viewed as an element of $$K = \operatorname{Sym}(G)$$, is an odd permutation. In other words, $$g \notin L$$: The cycle decomposition consists of $$m$$ cycles of length two each, i.e., an odd number of cycles of even length. Thus, $$g$$ is an odd permutation.
 * 3) $$G \cap L$$ is a subgroup of index two in $$G$$: By fact (3), $$G \cap L$$ has index either one or two in $$G$$. However, the previous step shows that $$G$$ is not contained in $$L$$, so $$G \cap L$$ is a proper subgroup of $$G$$. Thus, $$G \cap L$$ is a subgroup of index two in $$G$$.

Advanced proof
The statement is a particular case of the fact that cyclic Sylow subgroup for least prime divisor has normal complement, which in turn follows from Burnside's normal p-complement theorem.