Subgroups & Lagrange

|H| divides |G|; index [G:H] = |G|/|H|.

The idea

Theorem (Lagrange's theorem).

If $G$ is a finite group and $H$ is a subgroup of $G$, then $|H|$ divides $|G|$.

This is the first real constraint in group theory, and it is a strong one. It says a group of order $12$ has no subgroup of order $5$ whatever else is true of it — and no element of order $5$ either, since the powers of an element form a subgroup.

The quotient has a meaning of its own. For each $g \in G$ the translate $gH = \{gh : h \in H\}$ is called a coset of $H$, and the number of distinct cosets is the index $[G : H]$. The theorem in its counting form reads $|G| = [G : H] \cdot |H|.$

Notice what the theorem does not claim. It constrains which orders are possible; it does not promise they occur. A divisor of $|G|$ need not be the order of any subgroup — the alternating group $A_{4}$ has order $12$ and no subgroup of order $6$.

Ways to work on it

• Walkthrough. Lagrange's theorem, index, and element orders.
• Proof. See why — cosets partition G into equal pieces.
• Practice. Compute the index of a subgroup.
• Hardest. Prove prime-order groups are cyclic.

Not sure where to start? Take the ten-question placement test.