# Subgroup of index two is normal

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., subgroup of index two) must also satisfy the second subgroup property (i.e., normal subgroup)
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## Statement

### 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.

## Related facts

### 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. Poincare's theorem: This states that a subgroup of index $n$ contains a normal subgroup of index dividing $n!$.

## Proof

### 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.

## References

### Textbook references

• Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, More info, Page 74, Exercise 10(a) of Section 6 (Cosets)