Subgroups of Prime Index

The index multiplication formula has a sharp consequence when the index is prime. If a subgroup

H

has prime index in a finite group

G

, then no subgroup can sit strictly between

H

and

G

. This is a structural application of Lagrange's theorem: the index from

H

G

cannot split into two nontrivial positive factors. In this lecture, we prove the result and use it to identify when a subgroup is maximal with respect to inclusion.

Let

(G,\circ)

be a finite group and let

H

be a subgroup of

G

. The subgroup

H

is called a

\textbf{subgroup of prime index}

\begin{equation*} [G:H]=p \end{equation*}

for some prime number

p

The index multiplication formula has a sharp consequence when the index is prime. If a subgroup

H

has prime index in a finite group

G

, then no subgroup can sit strictly between

H

and

G

. This is a structural application of Lagrange's theorem: the index from

H

G

cannot split into two nontrivial positive factors. In this lecture, we prove the result and use it to identify when a subgroup is maximal with respect to inclusion.

Let

(G,\circ)

be a finite group and let

H

be a subgroup of

G

. The subgroup

H

is called a

\textbf{subgroup of prime index}

\begin{equation*} [G:H]=p \end{equation*}

for some prime number

p

Let

(G,\circ)

be a group and let

H

be a proper subgroup of

G

. The subgroup

H

is called a

\textbf{maximal subgroup}

G

if there is no subgroup

K

G

such that

\begin{equation*} H<K<G. \end{equation*}

Let

(G,\circ)

be a finite group and let

H

be a subgroup of

G

. If

[G:H]

is prime, then

H

is a maximal subgroup of

G

Given that

(G,\circ)

is a finite group,

H

is a subgroup of

G

, and

[G:H]

is prime. To prove that

H

is a maximal subgroup of

G

. Let

\begin{equation*} [G:H]=p, \end{equation*}

where

p

is prime. Let

K

be a subgroup of

G

such that

\begin{equation*} H\leq K\leq G. \end{equation*}

By the index multiplication formula,

\begin{equation*} [G:H]=[G:K][K:H]. \end{equation*}

Therefore

\begin{equation*} p=[G:K][K:H]. \end{equation*}

Since

p

is prime, either

\begin{equation*} [G:K]=1 \end{equation*}

\begin{equation*} [K:H]=1. \end{equation*}

[G:K]=1

, then

K=G

. If

[K:H]=1

, then

K=H

. Therefore every subgroup

K

satisfying

H\leq K\leq G

is equal to

H

G

. Hence,

H

is a maximal subgroup of

G

\square

We observe the interval of possible subgroups between

H

and

G

through the factorization of

[G:H]

. Enter an index and compare the available factorizations of that index. When the index is prime, there is no way to split it into two factors both larger than

1

, so an intermediate subgroup cannot appear through the index multiplication formula. When the index is composite, the theorem does not decide maximality, but the factorization shows why the prime argument no longer works.

The absence of nontrivial factorization is the key point. Prime index forces every subgroup between

H

and

G

to collapse to one endpoint, which is exactly maximality.

Let

(G,\circ)

be a finite group and let

H

be a subgroup of

G

. If

\begin{equation*} |G|=p|H| \end{equation*}

for some prime number

p

, then

H

is a maximal subgroup of

G

Given that

(G,\circ)

is a finite group,

H

is a subgroup of

G

, and

|G|=p|H|

for some prime number

p

. To prove that

H

is a maximal subgroup of

G

. By Lagrange's theorem,

\begin{align*} [G:H] &=\frac{|G|}{|H|}\\ &=\frac{p|H|}{|H|}\\ &=p. \end{align*}

Since

p

is prime,

[G:H]

is prime. By the previous theorem,

H

is a maximal subgroup of

G

. Hence,

H

is a maximal subgroup of

G

\square

Let

G=\mathbb{Z}_6

under addition modulo

6

and let

\begin{equation*} H=\{0,3\}. \end{equation*}

Then

\begin{equation*} |G|=6 \end{equation*}

and

\begin{equation*} |H|=2. \end{equation*}

Therefore

\begin{align*} [G:H] &=\frac{6}{2}\\ &=3. \end{align*}

Since

3

is prime,

H

is a maximal subgroup of

\mathbb{Z}_6

Let

G=\mathbb{Z}_{12}

and let

\begin{equation*} H=\{0,3,6,9\}. \end{equation*}

Then

|H|=4

and

\begin{equation*} [G:H]=\frac{12}{4}=3. \end{equation*}

Thus

H

has prime index, but

H

does not have prime order. Prime index is a statement about the number of cosets, not about the number of elements inside the subgroup.

Let

(G,\circ)

be a finite group with

|G|=30

. Let

H

be a subgroup of

G

with

|H|=10

. Prove that

H

is maximal.

Let

(G,\circ)

be a finite group with

|G|=30

and let

H

be a subgroup of

G

with

|H|=10

. By Lagrange's theorem,

\begin{align*} [G:H] &=\frac{|G|}{|H|}\\ &=\frac{30}{10}\\ &=3. \end{align*}

Since

3

is prime,

H

is a maximal subgroup of

G

We compute the index

[G:H]

from the orders of

G

and

H

. If the index is prime, the theorem proves that

H

is maximal in

G

. If the index is not prime, the calculator says only that this theorem does not apply. This avoids the common mistake of treating a failed sufficient condition as proof that

H

is not maximal.

1. Let

G

be a finite group with

|G|=42

and let

H

be a subgroup with

|H|=14

. Prove that

H

is maximal. 2. Let

G

be a finite group with

|G|=48

and let

H

be a subgroup with

|H|=12

. Does the prime-index theorem prove that

H

is maximal? 3. Let

H\leq K\leq G

and suppose

[G:H]=5

. Prove that

K=H

K=G

1. We have

\begin{equation*} [G:H]=\frac{42}{14}=3. \end{equation*}

Since

3

is prime,

H

is maximal. 2. We have

\begin{equation*} [G:H]=\frac{48}{12}=4. \end{equation*}

Since

4

is not prime, the prime-index theorem does not apply. This does not decide maximality. 3. By the index multiplication formula,

\begin{equation*} [G:H]=[G:K][K:H]. \end{equation*}

Thus

\begin{equation*} 5=[G:K][K:H]. \end{equation*}

Since

5

is prime, either

[G:K]=1

[K:H]=1

. Hence,

K=G

K=H

It means the number of cosets of

H

G

is a prime number.

An intermediate subgroup would split the prime index into two positive factors, which is impossible except by using

1

Not in general. The theorem in this lecture gives one sufficient condition for maximality.

BMLabs Mathematics

BMLabs Mathematics

Subgroups of Prime Index

Subgroups of Prime Index

OVERVIEW

SUBGROUP OF PRIME INDEX

MAXIMAL SUBGROUP

EXPLORE PRIME INDEX MAXIMALITY

PRIME INDEX MAXIMALITY EXPLORER

WHAT PRIME INDEX RULES OUT

A MAXIMAL SUBGROUP IN INTEGERS MODULO SIX

PRIME INDEX DOES NOT MEAN PRIME ORDER

TESTING MAXIMALITY BY INDEX

SOLUTION

TEST MAXIMALITY BY INDEX

PRIME INDEX MAXIMALITY CALCULATOR

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation