# Subsubgroups are subgroups of subgroups / Multiplicative Property of the Index

Algebra by Michael Artin Prop 2.8.14 Multiplicative Property of the Index

Statement of Prop 2.8.14

Let $G \supseteq H \supseteq K$ be subgroups of a group G. Then $[G:K] = [G:H][H:K]$.

Proof of Prop 2.8.14 Here's the original Counting Formula (Formula 2.8.8) Question: Could we perhaps prove Prop 2.8.14 by using Counting Formula 2.8.8? I think we could do so (Tower Law for Subgroups) if we show that $K$ is a subgroup of $H$ which I guess hasn't been proven yet, if it's even true.

The folklore is true: $$K$$, a subset of a subgroup $$H$$ of group $$G$$ and a subgroup of $$G$$, is a subgroup of $$H$$!

One hint we know it's true is that the above linked Tower Law for Subgroups (the difference there is that $$K$$ is a subgroup of $$H$$ is assumed and that the folklore there is that $$K$$ is a subgroup of $$G$$) has similar proof to Artin for Prop 2.8.14 (and these: 1, 2, 3).

Proof that $$K$$, a subset of a subgroup $$H$$ of group $$G$$ and a subgroup of $$G$$, is a subgroup of $$H$$:

1. Subset: $$K \subseteq H$$ by assumption.

2. Closure: Let $$k_1,k_2 \in K$$. Because $$K \subseteq G$$ is a subgroup of $$G$$, $$k_1k_2 \in K$$, which is the same requirement of closure for $$K \subseteq H$$ to be a subgroup of $$H$$.

3. Existence of Identity: Because $$K \subseteq G$$ is a subgroup of $$G$$, $$K$$ has an identity $$1_K$$, and, by Exer 2.2.5, $$1_K$$ is the identity of $$1_G$$, i.e. $$1_K=1_G$$. Because $$H \subseteq G$$ is a subgroup of $$G$$, $$H$$ has an identity $$1_H$$, and, by Exer 2.2.5, $$1_H$$ is the identity of $$1_G$$, i.e. $$1_H=1_G$$. Therefore, $$1_K=1_H$$, i.e. $$K$$ has an identity, and it is the identity in $$H$$.

4. Existence of Inverse: Let $$k_1 \in K$$. Because $$K \subseteq G$$ is a subgroup of $$G$$, there exists a $$k_3 \in K$$ s.t. $$k_3k_1=k_1k_3=1$$, which is the same requirement of existence of inverses for $$K \subseteq H$$ to be a subgroup of $$H$$.

QED

Note: Unlike in the analogues below, we used that $$H$$ is a subgroup of $$G$$.

I'm open other proofs that do not use that $$H$$ is a subgroup of $$G$$.

Proof of Prop 2.8.14:

By twice application of Counting Formula (Formula 2.8.8) with what we just proved, we have that for finite orders

$$[G:K] = \frac{|G|}{|K|}, [H:K] = \frac{|H|}{|K|}, [G:H] = \frac{|G|}{|H|}$$

Therefore, the result follows.

For infinite orders, it looks like we'll have to use the kind of proof with listing the cosets.

QED