# Tower Law for Subgroups

## Theorem

Let $\struct {G, \circ}$ be a group.

Let $H$ be a subgroup of $G$ with finite index.

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

Then:

$\index G K = \index G H \index H K$

where $\index G H$ denotes the index of $H$ in $G$.

## Proof 1

Let $p = \index G H$, $q = \index H K$.

By hypothesis these numbers are finite.

Therefore, there exist $g_1, \ldots, g_p \in G$ such that $G$ is a disjoint union: $\ds G = \bigsqcup_{i \mathop = 1}^p g_i H$

Similarly, there exist $h_1,\ldots,h_q \in H$ such that $H$ is a disjoint union: $\ds H = \bigsqcup_{j \mathop = 1}^q h_j K$

Thus:

 $\ds G$ $=$ $\ds \bigsqcup_{i \mathop = 1}^p g_i H$ $\ds$ $=$ $\ds \bigsqcup_{i \mathop = 1}^p g_i \bigsqcup_{j \mathop = 1}^q h_j K$ $\ds$ $=$ $\ds \bigsqcup_{i \mathop = 1}^p \bigsqcup_{j \mathop = 1}^q g_i \paren {h_j K}$ Product of Subset with Union $\ds$ $=$ $\ds \bigsqcup_{i \mathop = 1}^p \bigsqcup_{j \mathop = 1}^q \paren {g_i h_j} K$ Subset Product within Semigroup is Associative: Corollary

This expression for $G$ is the disjoint union of $p q$ cosets.

Therefore the number of elements of the coset space is:

$\index G K = p q = \index G H \index H K$

$\blacksquare$

## Proof 2

Assume $G$ is finite.

Then:

 $\ds \index G H$ $=$ $\ds \frac {\order G} {\order H}$ Lagrange's Theorem $\ds \index G K$ $=$ $\ds \frac {\order G} {\order K}$ Lagrange's Theorem $\ds \leadsto \ \$ $\ds \index G K$ $=$ $\ds \frac {\order H} {\order K} \times \index G H$

Since $K \le H$, from Lagrange's Theorem we have that $\dfrac {\order H} {\order K} = \index H K$.

Hence the result.

$\blacksquare$