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Problem 86 (Lagrange's Theorem Proof)

Prove Lagrange's Theorem.

Theorem (Lagrange's Theorem)

Suppose that $G$ is a finite group and that $H$ is a subgroup of $G$. Then the order of $H$ divides the order of $G$. In particular, we know that $|G|/|H|$ equals the number of distinct right (or left) cosets of $H$ in $G$.

Solution

Let $S = \left\{ Ha | a \in G \right\}$ and we let $r = |S|$. Since we know that either $Ha = Hb$ or $Ha \cap Hb = \emptyset$, for any $a, b \in G$ it follows that $S$ must be the set of distinct cosets of $H$, i.e. $S = \{ Ha_1, Ha_2, ..., Ha_r \}$. We also know that for every $g \in G$ we have $g \in Hg$, and therefore we may rewrite $G$ as the union of the distinct cosets of $H$, namely $G = \cup_{i=1}^r Ha_i$. To find the cardinality of $G$, recall that the cardinality of any intersection of a pair of cosets is empty and further that $|Ha| = |H|$ for any $a \in G$. We now compute

$$ \begin{align} |G| &= |\cup_{i=1}^r Ha_i| \\ &= \sum_{i=1}^r |Ha_i| \\ &= \sum_{i=1}^r |H| \\ &= r \cdot |H|. \end{align} $$

Q.E.D.

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