Lagrange's Theorem

Since the cosets of $H$ are equal-size pieces that fill up $G$, counting them gives one of the most useful theorems in algebra.

Why it works

Lagrange's theorem. If $G$ is finite and $H \le G$, then $|H|$ divides $|G|$, and the number of left cosets is $$[G : H] = \frac{|G|}{|H|}.$$ Proof. The left cosets of $H$ are disjoint and cover $G$, and each has exactly $|H|$ elements (the map $h \mapsto ah$ is a bijection $H \to aH$). If there are $k$ of them, then $|G| = k\,|H|$. So $|H|$ divides $|G|$, and $k = |G|/|H|$. $\square$

The number $[G : H]$ is the index of $H$ in $G$. Indices multiply along a chain $K \le H \le G$: $[G : K] = [G : H]\,[H : K]$, since $\frac{|G|}{|K|} = \frac{|G|}{|H|} \cdot \frac{|H|}{|K|}$.

Look at the shape of that proof: a partition into equal pieces turns a structure question into arithmetic. You will see the same move again in the orbit–stabilizer theorem.

Worked example: Possible subgroup orders

A group has order $12$. Which orders can its subgroups have? What about a group of order $15$?

By Lagrange, subgroup orders divide $12$: $1, 2, 3, 4, 6, 12$.
For order $15$: $1, 3, 5, 15$. A group of order $15$ has no subgroup of order $2$, $4$ or $6$.

Answer: Divisors of $|G|$ only.

Worked example: Intersections

$H$ and $K$ are subgroups of a group, with $|H| = 8$ and $|K| = 15$. What is $H \cap K$?

$H \cap K$ is a subgroup of both, so its order divides $8$ and $15$.
$\gcd(8, 15) = 1$, so $|H \cap K| = 1$.

Answer: $H \cap K = \{e\}$.

Worked example: Index in a chain

$K \le H \le G$ with $|G| = 60$, $[G : H] = 5$ and $|K| = 3$. Find $[H : K]$.

$|H| = 60/5 = 12$.
$[H : K] = 12/3 = 4$. Check: $[G : K] = 60/3 = 20 = 5 \cdot 4$.

Answer: $[H : K] = 4$.

The converse is false

Lagrange says subgroup orders must divide $|G|$. It does not say every divisor occurs. The classic example is $A_{4}$, of order $12$, which has no subgroup of order $6$. A subgroup of order $6$ would have index $2$; you will see that such a subgroup contains the square of every element. In $A_{4}$ the squares of the eight 3-cycles are again the eight 3-cycles (since $(a\;b\;c)^{2} = (a\;c\;b)$), too many for a group of order $6$.

The ten subgroups of $A_{4}$: orders 1, 2, 3, 4 and 12 occur, but no subgroup has order 6, even though 6 divides 12.

Common mistake

Believing the converse. "6 divides 12, so a group of order 12 has a subgroup of order 6" is false for $A_{4}$. Lagrange's theorem only rules out orders. (Partial converses exist: for cyclic groups every divisor occurs, and the Sylow theorems later guarantee prime-power orders.)

Key idea

Lagrange: for finite $G$, the order of every subgroup divides $|G|$, and there are $[G : H] = |G|/|H|$ cosets. The converse fails.

Sources and further reading

Thomas W. Judson, Abstract Algebra: Theory and Applications (open textbook), chapter 6, Cosets and Lagrange's Theorem (section 6.2, Lagrange's Theorem)

MIT OpenCourseWare 18.703 Modern Algebra (Spring 2013), lecture on cosets and Lagrange

John A. Beachy and William D. Blair, Abstract Algebra, online study guide (Northern Illinois University), section 3.2 (Lagrange's theorem)

Lagrange's Theorem

The converse is false

Practice

Builds on

Leads to

Sources and further reading