Tower Law for Subgroups

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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: $\displaystyle G = \bigsqcup_{i \mathop = 1}^p g_i H$

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

Thus:

\(\displaystyle G\) \(=\) \(\displaystyle \bigsqcup_{i \mathop = 1}^p g_i H\)
\(\displaystyle \) \(=\) \(\displaystyle \bigsqcup_{i \mathop = 1}^p g_i \bigsqcup_{j \mathop = 1}^q h_j K\)
\(\displaystyle \) \(=\) \(\displaystyle \bigsqcup_{i \mathop = 1}^p \bigsqcup_{j \mathop = 1}^q g_i \paren {h_j K}\) Product of Subset with Union
\(\displaystyle \) \(=\) \(\displaystyle \bigsqcup_{i \mathop = 1}^p \bigsqcup_{j \mathop = 1}^q \paren {g_i h_j} K\) 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:

\(\displaystyle \index G H\) \(=\) \(\displaystyle \frac {\order G} {\order H}\) Lagrange's Theorem
\(\displaystyle \index G K\) \(=\) \(\displaystyle \frac {\order G} {\order K}\) Lagrange's Theorem
\(\displaystyle \leadsto \ \ \) \(\displaystyle \index G K\) \(=\) \(\displaystyle \frac {\order H} {\order K} \times \index G H\)


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

Hence the result.

$\blacksquare$


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