# 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:

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

This needs considerable tedious hard slog to complete it.In particular: This does not get us where we wantTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

Due to the organization of pages at $\mathsf{Pr} \infty \mathsf{fWiki}$, this argument is circular.In particular: This is Index in Subgroup, which is used to prove the main theoremYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving this issue.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{CircularStructure}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

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