# Index of a subgroup

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## Definition

### Symbol-free definition

The index of a subgroup in a group is the following equivalent things:

1. The number of left cosets of the subgroup
2. The number of right cosets of the subgroup

The collection of left cosets is sometimes termed the coset space, so in this language, the index of a subgroup is the cardinality of its coset space.

### Definition with symbols

Given a subgroup $H$ of a group $G$, the index of $H$ in $G$, denoted $[G:H]$, is defined in the following ways:

1. It is the number of left cosets of $H$ in $G$, i.e. the number of sets of the form $xH$.
2. It is the number of right cosets of $H$ in $G$, i.e. the number of sets of the form $Hx$.

The collection of left cosets of $H$ in $G$ is sometimes termed the coset space, and is denoted $G/H$. With this notation, the index of $H$ in $G$, is the cardinality $\left|G/H\right|$.

### Equivalence of definitions

The equivalence of definitions follows from the fact that there is a natural bijection between the collection of left cosets of a subgroup, and the collection of its right cosets, given by the map $g \mapsto g^{-1}$

Further information: Left and right coset spaces are naturally isomorphic

### Further note for finite groups

When the group is finite, then by Lagrange's theorem, the index of a subgroup is the ratio of the order of the group to the order of the subgroup.

## Facts

### Multiplicativity of the index

Further information: Index is multiplicative If $H \le K \le G$, then we have:

$[G:K][K:H] = [G:H]$

In other words, the number of cosets of $H$ in $G$ equals the number of cosets of $H$ in $K$, times the number of cosets of $K$ in $G$.

In fact, more is true. We can set up a bijection as follows:

$G/K \times K/H \to G/H$

However, this bijection is not a natural one, and, in order to define it, we first need to choose a system of coset representatives of $H$.

### Effect of intersection on the index

Further information: Conjugate-intersection index theorem If $H_1$ and $H_2$ are two subgroups of $G$, then the index of $H_1 \cap H_2$ is bounded above by the product of the indices of $H_1$ and of $H_2$.

This follows as a consequence of the product formula. Note that equality holds if and only if $H_1H_2 = G$.

Note that in case $H_1$ and $H_2$ are conjugate subgroups of index $r$, the index of $H_1 \cap H_2$ is bounded above by $r(r-1)$.

## References

### Textbook references

• Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, More info, Page 57, Point (6.8) (definition in paragraph, defined as number of left cosets)
• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 90 (formal definition, defined as number of left cosets)