# Index of Intersection of Subgroups/Corollary

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## Theorem

Let $G$ be a group.

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

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

Then:

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

where $\index G K$ denotes the index of $K$ 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 $H K = \set {h k: h \in H, k \in K} = G$.

## Proof

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

From Index of Intersection of Subgroups, we have:

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

Setting $G = H$, we have:

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

$\blacksquare$

## Sources

- 1967: John D. Dixon:
*Problems in Group Theory*... (previous) ... (next): $1$: Subgroups: $1.\text{T}.2$