Subgroup of Index 2 is Normal
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Theorem
A subgroup of index $2$ is always normal.
Proof
Suppose $H \le G$ such that $\index G H = 2$.
Thus $H$ has two left cosets (and two right cosets) in $G$.
If $g \in H$, then $g H = H = H g$.
If $g \notin H$, then $g H = G \setminus H$ as there are only two cosets and the cosets partition $G$.
For the same reason, $g \notin H \implies H g = G \setminus H$.
That is, $g H = H g$.
The result follows from the definition of normal subgroup.
$\blacksquare$
Also see
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.9$
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.10$: Theorem $23$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \gamma$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 49.2$ Normal subgroups
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Example $7.6$