Subgroup of index two is normal

Verbal statement
Any subgroup whose index is two, is a normal subgroup.

Property-theoretic statement
The property of being a subgroup whose index in the whole group is two, is stronger than the property of being a normal subgroup.

General examples

 * In a symmetric group on a finite set, the alternating group is a subgroup of index two that is normal. It is also the kernel of the sign homomorphism. More generally, in a finitary symmetric group, the corresponding finitary alternating group is normal.
 * In a dihedral group, the corresponding cyclic maximal subgroup is normal. More generally, in a generalized dihedral group, the corresponding abelian maximal subgroup is normal.

Specific examples

 * In dihedral group:D8, the two Klein four-subgroups have index two and are normal. The cyclic maximal subgroup also has index two and is normal.
 * In quaternion group, the three cyclic maximal subgroups have index two and are normal.
 * In symmetric group:S3, the alternating group which is generated by a $$3$$-cycle has index two and in normal. Similarly, A4 in S4 is normal and A5 in S5 is normal.

Similar facts in group theory

 * Every group is normal in itself
 * Subgroup of index equal to least prime divisor of group order is normal: This states that a subgroup whose index is the least prime divisor of the order of the group is a normal subgroup.
 * Prime power order implies nilpotent, nilpotent implies every maximal subgroup is normal: In particular, in a nilpotent group, a subgroup of prime index is normal.
 * Poincare's theorem: Any subgroup of finite index contains a normal subgroup of finite index.
 * Union of three subgroups is the whole group implies they have index two and form a flower arrangement

Dual facts

 * Normal of order two implies central
 * Normal of order equal to least prime divisor of group order implies central

Opposite facts in group theory

 * Index two not implies characteristic
 * Index two of index two not implies normal: A subgroup of index two inside a subgroup of index two need not be normal.

Similar statements in relation to quasigroups and algebra loops

 * Subloop of index two is normal
 * Subquasigroup of size more than half is whole quasigroup
 * Subgroup of size more than half is whole group

Breakdown for Lie rings

 * Lie subring of index two not is ideal: The analogue of subgroups for a Lie ring is Lie subrings, and the analogue of normal subgroup is ideal. Thus, we might expect that a Lie subring of index two is an ideal. This, however, is not true.

Related facts in other disciplines

 * Any quadratic extension is normal: This is directly related, because an extension is normal if and only if the absolute Galois group of the larger field is normal in the absolute Galois group of the smaller field. Since the extension having degree two implies the index is two, this yields that any quadratic extension is normal. (The standard proof is more direct).
 * Any double cover is regular: A covering map where the fibers are of size two, i.e., a double cover, is a regular covering. Here, the subgroup of the fundamental group corresponding to the covering is an index two subgroup, hence normal.

Related numerical facts

 * $$2 - 1 = 1$$ (see the proof involving the coset formulation; also, the fact that factoring out one root of a quadratic polynomial gives a linear polynomial, which must therefore have a root in the same field).
 * $$2! = 2$$ (see the proof involving the group action on the left coset space; also, the Galois theory interpretation, which notes that the Galois group in this case is the full symmetric group).

Facts used

 * 1) uses::Poincare's theorem: This states that a subgroup of index $$n$$ contains a normal subgroup of index dividing $$n!$$.

Proof in terms of cosets definition of normality
The proof is direct if we use the following definition of normality: a subgroup $$H$$ is normal in $$G$$ if the left cosets and the right cosets of $$H$$ coincide.

Given: A group $$G$$, a subgroup $$H$$ of index two.

To prove: $$H$$ is normal in $$G$$.

Proof: Observe that since $$H$$ has index two, it has exactly two left cosets: $$H$$ and the set of elements $$G \setminus H$$. $$H$$ also has exactly two right cosets: $$H$$ and $$G \setminus H$$. Thus, the left and right cosets of $$H$$ coincide.

Proof using group action on the left coset space
Given: A group $$G$$, a subgroup $$H$$ of index two.

To prove: $$H$$ is normal in $$G$$.

Proof: By fact (1), $$H$$ contains a normal subgroup of $$G$$ of index dividing $$2!$$. But since $$2 = 2!$$, this normal subgroup must coincide with $$H$$.

Note that fact (1) in turn follows from the fact that a group acts on the left coset space of any subgroup by left multiplication.

Textbook references

 * , Page 74, Exercise 10(a) of Section 6 (Cosets)