# Index of Intersection of Subgroups

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## Contents

## Theorem

Let $G$ be a group.

Let $H, K$ be subgroups of finite index of $G$.

Then:

- $\index G {H \cap K} \le \index G H \index G K$

where $\index G H$ denotes the index of $H$ in $G$.

Note that here the symbol $\le$ is being used with its meaning **less than or equal to**.

Equality holds if and only if $H K = \set {h k: h \in H, k \in K} = G$.

### Corollary

Let $H$ be a subgroup of $G$.

Let $K$ be a subgroup of finite index of $G$.

Then:

- $\index H {H \cap K} \le \index G K$

## Proof

Note that $H \cap K$ is a subgroup of $H$.

From Tower Law for Subgroups, we have:

- $\index G {H \cap K} = \index G H \index H {H \cap K}$

From Index in Subgroup, also:

- $\index G {H \cap K} \le \index G K$

Combining these results yields the desired inequality.

Again from Index in Subgroup, it follows that:

- $\index H {H \cap K} = \index G K$

if and only if $H K = G$.

$\blacksquare$

## Sources

- 1967: John D. Dixon:
*Problems in Group Theory*... (previous) ... (next): $1$: Subgroups: $1.\text{T}.2$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 39 \beta$ - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 3.3$: Group actions and coset decompositions: Exercise $9$