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Topic 2 of 8

Subgroups & Lagrange's Theorem

Not every subset of a group is itself a group — but those that are (subgroups) reveal the internal structure of $G$. Lagrange's theorem, one of the first major results in group theory, constrains which orders subgroups can have. Its consequences reach into number theory, geometry, and the representation theory used in particle physics.

Key Concepts

Subgroup

A subset

H \subseteq G

that is itself a group under the same operation. Equivalently:

e \in H

H

is closed under

\cdot

, and

a \in H \Rightarrow a^{-1} \in H

Left Coset

For

g \in G

gH = \{gh : h \in H\}

. Cosets partition

G

into equal-sized disjoint subsets, each of size

|H|

Index

[G:H] = |G|/|H|

, the number of distinct left cosets. Also equals the number of right cosets.

Normal Subgroup

H \trianglelefteq G

gHg^{-1} = H

for all

g \in G

, i.e. left and right cosets coincide. Kernels of homomorphisms are always normal subgroups.

Quotient Group

When

H \trianglelefteq G

, the set of cosets

G/H

forms a group of order

|G|/|H|

. The Standard Model symmetry breaking

SU(2)\times U(1) \to U(1)_{\text{EM}}

is a quotient construction.

Key Equations

Lagrange's Theorem

|H| \,\Big|\, |G| \qquad \text{and} \qquad [G:H] = \frac{|G|}{|H|}

For any subgroup

H \leq G

(finite

G

), the order of

H

divides the order of

G

Normal Subgroup Condition

H \trianglelefteq G \iff gH = Hg \quad \forall\, g \in G

Equivalently:

ghg^{-1} \in H

for all

g \in G

h \in H

Worked Example

Index of a Subgroup of $\mathbb{Z}_{12}$

Problem

In $\mathbb{Z}_{12}$ , consider the subgroup $H = \{0, 4, 8\}$ . Find $|H|$ and the index $[\mathbb{Z}_{12} : H]$ .

Solution

$H$ has 3 elements, so $|H| = 3$ . Check it's a subgroup: $4+4=8$ , $4+8=12\equiv0$ , $8+8=16\equiv4$ . Closed ✓

By Lagrange: $[\mathbb{Z}_{12}:H] = |\mathbb{Z}_{12}|/|H| = 12/3 = 4$ .

The 4 cosets are: $0+H=\{0,4,8\}$ , $1+H=\{1,5,9\}$ , $2+H=\{2,6,10\}$ , $3+H=\{3,7,11\}$ . They partition $\mathbb{Z}_{12}$ . ✓

Answer

|H| = 3

\;[\mathbb{Z}_{12}:H] = 4

Practice

Exercises

6 problems

1 of 6

If $|G| = 36$ and $H \leq G$ with $|H| = 9$ , what is the index $[G:H]$ ?

Your answer

2 of 6

$\mathbb{Z}_{24}$ has exactly one subgroup for each divisor of $24$ . How many subgroups does $\mathbb{Z}_{24}$ have?

Your answer

3 of 6

In $\mathbb{Z}_{15}$ , what is the order of the cyclic subgroup $H = \langle 3 \rangle = \{0, 3, 6, 9, 12\}$ ?

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4 of 6

$S_4$ has order 24. The alternating group $A_4 \leq S_4$ has order 12. What is $[S_4 : A_4]$ ?

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5 of 6

A group $G$ has $|G| = 42$ . By Lagrange's theorem, subgroup orders must divide 42. How many positive divisors does 42 have? (These are all the possible subgroup orders.)

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6 of 6

In $\mathbb{Z}_{60}$ , the subgroup $H = \langle 12 \rangle$ has order $60/\gcd(60,12) = 5$ . What is the index $[\mathbb{Z}_{60} : H]$ ?

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Key Takeaways

A subgroup $H \leq G$ must contain the identity, be closed under the operation, and contain the inverse of every element.
Lagrange's theorem: $|H|$ divides $|G|$ . Consequently, the order of every element divides $|G|$ .
The index $[G:H] = |G|/|H|$ counts how many cosets partition $G$ . Cosets are either identical or disjoint.
Normal subgroups ( $H \trianglelefteq G$ ) allow formation of the quotient group $G/H$ . All subgroups of abelian groups are normal.

Subgroups & Lagrange's Theorem

Key Concepts

Key Equations

Index of a Subgroup of Z12\mathbb{Z}_{12}Z12​

Exercises

Key Takeaways

Index of a Subgroup of $\mathbb{Z}_{12}$