Index is One iff Subgroup equals Group
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Theorem
Let $G$ be a group whose identity element is $e$.
Let $H$ be a subgroup of $G$.
Then:
- $\index G H = 1 \iff G = H$
where $\index G H$ denotes the index of $H$ in $G$.
Proof
For finite groups, we can apply Lagrange's Theorem:
- $\index G H = \dfrac {\order G} {\order H}$
But then:
- $\dfrac {\order G} {\order H} = 1 \iff \order G = \order H$
Hence the result.
For the general case (including infinite groups) we need to consider the (left) coset space $G / H$.
Note that we must have $e H = H \in G / H$.
Hence:
- $\index G H = 1 \iff G / H = \set H \iff G = H$
$\blacksquare$
Sources
- 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts