Tower Law for Subgroups/Proof 1

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

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

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


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

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


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: $\ds H = \bigsqcup_{j \mathop = 1}^q h_j K$


\(\ds G\) \(=\) \(\ds \bigsqcup_{i \mathop = 1}^p g_i H\)
\(\ds \) \(=\) \(\ds \bigsqcup_{i \mathop = 1}^p g_i \bigsqcup_{j \mathop = 1}^q h_j K\)
\(\ds \) \(=\) \(\ds \bigsqcup_{i \mathop = 1}^p \bigsqcup_{j \mathop = 1}^q g_i \paren {h_j K}\) 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:

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