# Difference between revisions of "Index is multiplicative"

## Statement

### Set-theoretic version

Suppose $K \le H \le G$ are groups. Then, we have a natural surjective map between the left coset spaces: $G/K \to G/H$

with the property that the inverse image of each point has size equals to the size of the left coset space $H/K$.

### Numerical version

Suppose $K \le H \le G$ are groups such that the indices $[G:H]$ and $[H:K]$ are finite. Then, we have: $[G:K] = [G:H][H:K]$.

Note that the statement makes sense even for infinite groups, if we interpret the cardinalities of the coset spaces as infinite cardinals and do the multiplication accordingly.

## Related facts

• Lagrange's theorem: Lagrange's theorem is a special case of this where $K$ is the trivial group.
• Third isomorphism theorem: The third isomorphism theorem is a stronger version of the statement where both subgroups are normal in the whole group. In this case, the surjective map from $G/K$ to $G/H$ is a homomorphism and the kernel is $H/K$.

## Facts used

1. Subgroup containment implies coset containment: If $K \le H \le G$ ,then every left coset of $K$ is contained in a unique left coset of $H$.

## Proof

### Proof of the set-theoretic version

Given: Groups $K \le H \le G$.

To prove: There is a surjective map from $G/K$ to $G/H$ where the inverse image of every point has size equal to the size of $H/K$.

Proof: Define $\varphi$ as the map sending a coset of $K$ to the unique coset of $H$ containing it (fact (1)). In other words: $\varphi(gK) = gH$. $\varphi$ is a well-defined map from $G/K$ to $G/H$. Further:

• $\varphi$ is surjective, since for any coset $gH$ of $H$ in $G$, $gH = \varphi(gK)$.
• The size of each inverse image equals the size of $H/K$: Consider a coset $gK$. We want to find all the left cosets of $H$ which map to this. This is equivalent to finding all the left cosets of $H$ contained in $gK$.

For this, consider a map $\psi$ that sends a left coset $xH$ of $H$ in $gK$, to the left coset $g^{-1}xH$.

Note that:

• This map $\psi$ is well-defined, because if $x$ is replaced by $xh, h \in H$, the left cosets $g^{-1}xH$ and $g^{-1}xhH$ are the same.
• The map $\psi$ sends left cosets of $H$ in $gK$ to left cosets of $H$ in $K$, because since $x \in gK$, $g^{-1}x \in K$.
• The map is injective, because it comes from a left multiplication on cosets.
• The map is surjective, because any left coset $kH$ of $H$ in $K$ arises as the image of the left coset $gkH$, which lies in $gK$.

Thus, $\psi$ is a bijection between the left cosets of $H$ in $gK$ and the left cosets of $H$ in $K$. Thus, the number of left cosets of $H$ in $gK$ equals the size of the left coset space $K/H$.

### Proof of the numerical version

The set-theoretic version shows that $G/K$ is the disjoint union of $|G/H|$ sets, each of size $|H/K|$. This yields: $|G/K| = |G/H||H/K|$.

By the definition of index of a subgroup, this yields: $[G:K] = [G:H][H:K]$.