# 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 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 Brauer core (?) of the whole group.

## Related facts

### Stronger facts

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

### 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. 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. Cauchy's theorem: If a prime $p$ divides the order of a finite group $G$, then $G$ has an element of order $p$.
3. Index satisfies transfer inequality: If $A,B$ are subgroups of $C$, then $[A:A \cap B] \le [C:B]$.

## Proof

### 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 $G$.