Index of Intersection of Subgroups/Corollary

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:

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

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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 theorem You 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: $1.\text{T}.2$