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Counting a group by its cosets: the order of a subgroup divides the order of the group, the index is the quotient, and the converse is false.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to assemble the proof of Lagrange's theorem from the coset facts, compute the index of a subgroup, use divisibility to rule out subgroup and element orders without computing anything, say precisely what the theorem does not claim, and give the standard counterexample to its converse.
The last lesson established three things about the cosets of a subgroup: they all have the same size, they never partly overlap, and together they cover the group. Nothing else is needed. This lesson does the arithmetic those three facts make available, and the result is the most quoted theorem in finite group theory.
Lagrange's theorem: for a subgroup $H$ of a finite group $G$, $|G| = [G : H]\,|H|$, so $|H|$ divides $|G|$. The index $[G : H]$ is the number of cosets. The converse would be the claim that every divisor of $|G|$ is the order of some subgroup; it is false.
> Lagrange's theorem. If $H$ is a subgroup of a finite group $G$, then > $$|G| = [G : H]\,|H|.$$ > In particular $|H|$ divides $|G|$.
The proof is one sentence of arithmetic on top of the last lesson. $G$ is the disjoint union of the $[G : H]$ cosets of $H$; each has $|H|$ elements; so counting $G$ by cosets gives $[G : H]\,|H|$, and counting it directly gives $|G|$.
What makes the theorem powerful is how little it uses. Nothing about the operation appears beyond cancellation, so it constrains every group of a given size simultaneously, whatever its multiplication table.
The corollaries do most of the work.
And the theorem does not converse. $|H|$ dividing $|G|$ does not produce a subgroup. The alternating group on four letters has order $12$ and no subgroup of order $6$. Nor does the theorem count subgroups: a cyclic group has exactly one of each permitted order, while the symmetries of a square have three of order $4$.
Most of the theorems in this course are counting arguments wearing algebraic clothes. Cosets all have the same size, so a subgroup's order divides the group's; the remainders below a degree are a finite list, so a quotient by an irreducible polynomial is a finite field.
Another way: picture
A chessboard covered exactly by dominoes: because every domino covers two squares and none overlaps and none hangs off, the number of squares must be twice the number of dominoes. Nobody needs to know anything about chess to draw that conclusion — only that the tiles are equal and the cover is exact. Lagrange is that argument with $|H|$ in place of $2$.
Another way: steps
To use Lagrange on a concrete question: 1. Write down $|G|$. 2. To rule out a subgroup or an element order, ask whether the candidate number divides $|G|$; if not, it cannot occur. 3. To find an index, divide. 4. To conclude something exists, stop — the theorem does not do that, and a separate argument is needed.
Two consequences are worth stating as results, because they classify rather than merely restrict.
Every group of prime order is cyclic. Take $g \ne e$ in a group of order $p$. Its order divides $p$ and is not $1$, so it is $p$, and $\langle g \rangle$ already has $p$ elements: it is the whole group. So up to isomorphism there is exactly one group of each prime order. No assumption about the operation was made, which is the striking part — a single divisibility fact settles the entire structure.
Fermat and Euler are corollaries. The units modulo $n$ form a group of order $\phi(n)$, so $a^{\phi(n)} \equiv 1 \pmod n$ for every $a$ coprime to $n$; when $n = p$ is prime that reads $a^{p-1} \equiv 1$. Two theorems of number theory, proved by counting cosets.
A third consequence has a subtler flavour: a group of order $2m$ with $m$ odd always contains an element of order $2$. Pair each element with its inverse; the pairs use up an even number of elements, the identity is left over on its own, so at least one more element must be its own inverse. This uses the same counting habit as Lagrange without using Lagrange itself.
Where the theorem stops is equally worth knowing. It gives no subgroup of any order, it counts no subgroups, and it says nothing about which groups of a given order exist. Sylow's theorems, in Abstract algebra II, supply partial converses: for a prime power dividing $|G|$, a subgroup of that order does exist, and the number of them is constrained. The general converse remains false.
Reading the theorem backwards. $|H| \mid |G|$ is a necessary condition, never sufficient. "Six divides twelve, so there is a subgroup of order six" is the error, and the even permutations of four letters are the counterexample to have ready.
Thinking the index is a difference. $[G : H] = |G|/|H|$. A group of order $12$ with a subgroup of order $3$ has index $4$, not $9$.
Applying the theorem to a subset that is not a subgroup. The cosets of $H$ are not subgroups, and no theorem says their sizes divide anything except by being $|H|$.
Assuming uniqueness. Lagrange permits; it does not count. Several subgroups of the same order may exist, and in non-cyclic groups usually do.
Using it on infinite groups without care. The statement about dividing needs $|G|$ finite. The index can still be defined and finite when the group is not — the even integers have index $2$ in the integers — and the useful form there is $|G| = [G:H]|H|$ read as a statement about cardinalities.
A group has $15$ elements. Could an element have order $4$?
The question, with nothing else given.
The order of an element is the size of the cyclic subgroup it generates, so by Lagrange it divides $15$. The divisors are $1, 3, 5, 15$.
Divisibility does the work.
$4$ is not among them, so no. Nothing about the group's operation was needed — the answer holds for every group of order $15$ there is.
One fact, every such group.
Work in $U(7) = \{1, \ldots, 6\}$ under multiplication modulo $7$: a group of order $6$.
A group, so Lagrange applies.
Every element's order divides $6$, so $a^{6} \equiv 1 \pmod 7$ for every $a$ not divisible by $7$.
The corollary $g^{|G|} = e$.
Check $a = 3$: $3^{6} = 729 = 104 \times 7 + 1$. And $3$ in fact has order $6$, so no smaller exponent works for it.
Fermat's little theorem, for free.
Index: $20/5 = 4$, so the subgroup has four cosets.
Divide.
Element orders divide $20$: the possibilities are $1, 2, 4, 5, 10, 20$.
The divisors, and nothing else.
Note that nothing here promises an element of order $10$ exists — only that no other order can. That is the theorem's one direction, used twice.
Build the proof of Lagrange's theorem: the order of a subgroup divides the order of the group.
This task has no paper form; do it on a device.
A group has $14$ elements and a subgroup $H$ has $7$. What is the index of $H$?
Answer:
A group has $8$ elements. Give the index of a subgroup of each of these sizes.
| Its index | |
|---|---|
| a subgroup with 4 elements | |
| a subgroup with 2 elements | |
| a subgroup with 8 elements |
Match each subgroup to its index in the group it sits in.
| index $2$ | index $3$ | index $4$ | index $6$ | index $12$ | |
|---|---|---|---|---|---|
| The even permutations, inside all the permutations of four letters | |||||
| The multiples of $4$, inside the integers modulo $12$ | |||||
| The multiples of $3$, inside the integers modulo $12$ | |||||
| The trivial subgroup, inside the integers modulo $6$ |
Select every statement that Lagrange's theorem supports.
This task has no paper form; do it on a device.
The even permutations of four letters form a group of order $12$, and $6$ divides $12$. How many subgroups of order $6$ does it have?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A group has $24$ elements. Give the index of a subgroup of each of these sizes.
| Its index | |
|---|---|
| a subgroup with 6 elements | |
| a subgroup with 4 elements | |
| a subgroup with 24 elements |
You can prove Lagrange's theorem from the tiling and use it to rule out orders. Say in your own words why a divisor of the group's order need not be the order of any subgroup. Next: the corollaries, which turn one divisibility fact into Fermat, Euler and the classification of groups of prime order.
9. Your turn: a group has $20$ elements and a subgroup has $5$. What is the index, and what element orders are possible?, step 3