DG10 - Lagrange’s theorem | AQA A-Level Further Mathematics

A subgroup cannot have an arbitrary size inside a finite group. You will use Lagrange’s theorem to restrict subgroup orders and element periods, then separate those necessary restrictions from proof that a particular subset is a subgroup.

Subgroup orders in finite groups

Can a group with $15$ elements contain a subgroup with $6$ elements? The issue is whether the larger group can be divided into equal-sized blocks built from the subgroup.

Start with addition modulo $12$ , and the subgroup $H=\{0,4,8\}$ . Translating its elements by $0,1,2,3$ gives

\begin{array}{c|ccc} H&0&4&8\\ 1+H&1&5&9\\ 2+H&2&6&10\\ 3+H&3&7&11 \end{array}

These blocks contain every group element exactly once, with three elements per block. Thus $12=4\times3$ .

The same idea works for any subgroup $H$ of a finite group $G$ . For each $g\in G$ , form the block $gH=\{g*h:h\in H\}$ . Cancellation makes each block have exactly $|H|$ elements. Every element belongs to a block, since $g=g*e$ . Two blocks are either identical or disjoint: if $g_1*h_1=g_2*h_2$ , then $g_1=g_2*h_2*h_1^{-1}$ , so every $g_1*h$ is in $g_2H$ ; reversing the argument gives equality. The whole group therefore splits into an integer number of equal-sized blocks.

Lagrange’s theorem

If $H$ is a subgroup of a finite group $G$ , then

|H|\mid |G|

That is, the order of $H$ divides the order of $G$ .

Not in the formula booklet: learn it.

First list positive divisors. Then apply any extra words: proper removes $|G|$ , while non-trivial removes $1$ . These are candidate orders; finding a subgroup still needs evidence.

Worked example 1: possible proper-subgroup orders

Question. A group $G$ has order $24$ . State, with a reason, the possible orders of its proper subgroups according to Lagrange’s theorem.

Plan. List divisors of $24$ , then exclude the whole group.

\begin{aligned} \text{positive divisors of }24&:1,2,3,4,6,8,12,24\\ \text{proper-subgroup candidates}&:1,2,3,4,6,8,12 \end{aligned}

The trivial subgroup has order $1$ and is proper here.

Answer. $1,2,3,4,6,8,12$ : Lagrange’s theorem requires division of $24$ , and proper excludes $24$ .

Check. Dividing $24$ by each listed number gives an integer. The factor pairs $(1,24),(2,12),(3,8),(4,6)$ show that no divisor has been missed.

Keep the same group order and add just the condition non-trivial.

Worked example 2: imposing both restrictions

Question. A group $G$ has order $24$ . State the possible orders of its proper, non-trivial subgroups according to Lagrange’s theorem.

Plan. Start with the same divisors and exclude both boundary cases.

\{1,2,3,4,6,8,12,24\}\setminus\{1,24\} =\{2,3,4,6,8,12\}

Answer. $2,3,4,6,8,12$ .

Check. These are exactly the divisors $d$ satisfying $1<d<24$ . The strict inequalities independently encode non-trivial and proper.

The opening question now has a definite answer: $6$ does not divide $15$ , so a subgroup of order $6$ is impossible in a group of order $15$ . Lagrange’s theorem concerns finite group orders; it is not an instruction to divide one infinity by another.

Watch out

“Proper” alone does not remove order $1$ . State the divisibility reason as well as the list when a question asks for a reason.

Apply both restrictions to a new group order.

Element periods and consequences

Repeating one element also creates a subgroup. If $a$ has period $k$ , its distinct powers before repetition are $e,a,a^2,\ldots,a^{k-1}$ . Products add exponents modulo $k$ , inverses are powers too, and associativity is inherited. This cyclic subgroup has order $k$ , so Lagrange’s theorem applies to it.

Period consequences

In a finite group $G$ of order $N$ , every element $a$ has a finite period $k$ , and

k\mid N,\qquad a^N=e

More generally, if $a^m=e$ for a positive integer $m$ , then $k\mid m$ . Under an additive operation, read $a^m$ as $m$ repetitions, usually written $ma$ .

Not in the formula booklet: learn it.

Why does the period divide any return exponent? Write $m=qk+r$ , with $0\le r<k$ . Then $a^m=(a^k)^q a^r=a^r$ . If this is $e$ , minimality of $k$ forces $r=0$ . Conversely, every multiple of $k$ gives the identity. A return exponent need not be the least return exponent.

Worked example 3: restricting a period

Question. A group has order $18$ . An element $a$ satisfies $a^6=e$ and $a^3\ne e$ . Find the remaining possible periods of $a$ .

Plan. Use divisibility for equalities and eliminate periods incompatible with the inequality.

\begin{aligned} k\mid18\text{ and }k\mid6&\quad\Longrightarrow\quad k\in\{1,2,3,6\}\\ a^3\ne e&\quad\Longrightarrow\quad k\nmid3 \end{aligned}

Thus periods $1$ and $3$ are excluded.

Answer. The period is $2$ or $6$ ; the given information does not distinguish them.

Check. Both occur in addition modulo $18$ : $9$ has period $2$ , and $3$ has period $6$ . Six copies of either give $0$ , but three copies give $9$ , verifying that neither candidate can be removed.

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Discrete Mathematics

DG10 - Lagrange’s theorem

Subgroup orders in finite groups

Element periods and consequences

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