FP2 1.4 - Lagrange's theorem

FP2 1.4 - Lagrange's theorem

Use Lagrange's theorem to restrict subgroup and element orders (FP2 1.4). You will see why subgroup sizes divide a finite group size, separate necessary conditions from existence claims, and explain why a group of prime order must be cyclic.

Why a subgroup size divides the group size

Could a group with 1515 elements contain a subgroup with 66 elements? The obstacle is a counting one: a finite group can be partitioned into equally sized copies of any subgroup, with no leftovers.

You already know

A subgroup contains the original identity and is closed under the original operation and inverses. If aa has order dd, the generated subgroup ⟨a⟩\langle a\rangle has dd elements. Equal-sized disjoint sets have total size equal to the number of sets times their common size.

For a concrete picture, take addition modulo 1212 and H={0,4,8}H=\{0,4,8\}. Adding 0,1,2,30,1,2,3, respectively, to all members of HH produces four disjoint three-element sets.

[DIAGRAM: asset_name: Equal-sized copies of an additive subgroup; asset_slug: lagrange-equal-copies; caption: Four disjoint copies, each with three members, account for all twelve group elements.; description: Four rows list H={0,4,8},1+H={1,5,9},2+H={2,6,10},3+H={3,7,11} in addition modulo12. Each row contains three equally sized cells. A separate equation says four times three equals twelve. Dark ink labels and two restrained accents distinguish the original subgroup from its translated copies.]
Four disjoint copies, each with three members, account for all twelve group elements.

The other rows need not themselves be subgroups: 1+H1+H, for example, does not contain the identity 00. They are equal-sized copies used for counting.

A translated copy of a subgroup

Writing an operation multiplicatively, for a∈Ga\in G define

aH={ah:h∈H}aH=\{ah:h\in H\}

This is obtained by multiplying every member of HH on the left by the same aa. In an additive example, the corresponding notation is a+Ha+H.

Here is why the counting works in any finite group. Cancellation makes h↦ahh\mapsto ah one-to-one, so each aHaH has ∣H∣|H| members. If two copies overlap, write ah1=bh2ah_1=bh_2. Then

a=bh2h1−1a=bh_2h_1^{-1}

and every ahah lies in bHbH, since

ah=b(h2h1−1h),h2h1−1h∈Hah=b(h_2h_1^{-1}h),\qquad h_2h_1^{-1}h\in H

The copies have the same finite size, so this inclusion makes them equal. Thus distinct copies are disjoint. They cover GG, because each gg belongs to gHgH through g=geg=ge.

Lagrange's theorem

The notation H≤GH\leq G means that HH is a subgroup of GG. If GG is finite, then

∣H∣∣∣G∣|H|\mid|G|

Equivalently, for some positive integer kk,

∣G∣=k∣H∣|G|=k|H|

The number kk counts the distinct equal-sized copies of HH.

Not in the formula booklet: learn it.

In the opening question, 66 does not divide 1515, so such a subgroup cannot exist.

Key idea

A subgroup partitions a finite group into equal-sized translated copies; divisibility is forced by that partition.

The theorem restricts possible sizes. It does not promise a subgroup of every size dividing ∣G∣|G|.

Restrict subgroup and element orders

Apply the theorem first to a subgroup HH, then to the subgroup generated by an element aa. This gives restrictions on both kinds of order.

Element-order consequences

For an element aa of a finite group GG,

ord⁡(a)=∣⟨a⟩∣∣∣G∣\operatorname{ord}(a)=|\langle a\rangle|\mid|G|

Consequently,

a∣G∣=ea^{|G|}=e

Indeed, if ∣G∣=kd|G|=kd with d=ord⁡(a)d=\operatorname{ord}(a), then a∣G∣=(ad)k=ea^{|G|}=(a^d)^k=e.

These are necessary restrictions. A divisor of ∣G∣|G| is not automatically the order of some element or subgroup.

Not in the formula booklet: learn it.

A proper subgroup excludes the full size ∣G∣|G|. If the question also says nontrivial, exclude size 11 as well.

Worked example 1: the divisors of eighteen

Question. A group has order 1818. List the subgroup orders that Lagrange permits for a proper subgroup, and the element orders it permits. Do these lists prove that all such objects exist?

Plan. List positive divisors, then apply the proper-subgroup restriction only to the subgroup list.

18=2(32)factor the group sized∣18⟹d∈{1,2,3,6,9,18}all positive divisors∣H∣<18⟹∣H∣∈{1,2,3,6,9}exclude the whole group\begin{aligned} 18&=2(3^2)&&\text{factor the group size}\\ d\mid18&\Longrightarrow d\in\{1,2,3,6,9,18\} &&\text{all positive divisors}\\ |H|<18&\Longrightarrow |H|\in\{1,2,3,6,9\} &&\text{exclude the whole group} \end{aligned}

Why this step? An element may generate the whole group, so order 1818 remains permitted for elements even though the subgroup is required to be proper.

Answer. Proper-subgroup sizes permitted: 1,2,3,6,91,2,3,6,9. Element orders permitted: 1,2,3,6,9,181,2,3,6,9,18. Lagrange alone does not guarantee each one's existence.

Check. Each listed size divides 1818; a proposed subgroup size 44 is excluded because 18/418/4 is not an integer.

Change only the group order from 1818 to 2020.

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