Subgroups of Prime Power Order Groups

Lagrange's theorem becomes very restrictive when the order of a group is a power of a prime. If

|G|=p^n

, then every subgroup order must divide

p^n

. The only positive divisors of

p^n

are

1,p,p^2,\ldots,p^n

. Therefore every subgroup of such a group has prime-power order with the same prime

p

. This lecture records that consequence carefully and uses it to rule out impossible subgroup orders.

Let

(G,\circ)

be a finite group. The group

G

is called a

\textbf{prime power order group}

\begin{equation*} |G|=p^n \end{equation*}

for some prime number

p

and some positive integer

n

Lagrange's theorem becomes very restrictive when the order of a group is a power of a prime. If

|G|=p^n

, then every subgroup order must divide

p^n

. The only positive divisors of

p^n

are

1,p,p^2,\ldots,p^n

. Therefore every subgroup of such a group has prime-power order with the same prime

p

. This lecture records that consequence carefully and uses it to rule out impossible subgroup orders.

Let

(G,\circ)

be a finite group. The group

G

is called a

\textbf{prime power order group}

\begin{equation*} |G|=p^n \end{equation*}

for some prime number

p

and some positive integer

n

Let

(G,\circ)

be a finite group. If

\begin{equation*} |G|=p^n, \end{equation*}

where

p

is prime and

n\in\mathbb{N}

, then every subgroup of

G

has order

p^m

for some integer

m

such that

\begin{equation*} 0\leq m\leq n. \end{equation*}

Given that

(G,\circ)

is a finite group and

\begin{equation*} |G|=p^n, \end{equation*}

where

p

is prime and

n\in\mathbb{N}

. To prove that every subgroup of

G

has order

p^m

for some integer

m

such that

0\leq m\leq n

. Let

H

be a subgroup of

G

. By Lagrange's theorem,

\begin{equation*} |H|\mid |G|. \end{equation*}

Therefore

\begin{equation*} |H|\mid p^n. \end{equation*}

Since

p

is prime, every positive divisor of

p^n

has the form

\begin{equation*} p^m \end{equation*}

where

m

is an integer satisfying

\begin{equation*} 0\leq m\leq n. \end{equation*}

Therefore

\begin{equation*} |H|=p^m \end{equation*}

for some integer

m

such that

0\leq m\leq n

. Hence, every subgroup of

G

has order

p^m

for some integer

m

such that

0\leq m\leq n

\square

We observe the possible subgroup orders when the group order is

p^n

. Choose a prime

p

and an exponent

n

, and the ladder displays the only orders that Lagrange's theorem allows. As

n

increases, the ladder grows by multiplying by the same prime each time. Try testing an order such as

6

when

|G|=32

and notice that it is ruled out because it does not divide the prime power.

The ladder lists divisors of the group order, not guaranteed subgroup orders. Lagrange's theorem is being used as a restriction: any order outside the ladder is impossible.

Let

(G,\circ)

be a finite group with

\begin{equation*} |G|=p^n, \end{equation*}

where

p

is prime. If

q

is a prime number different from

p

, then

G

has no subgroup of order divisible by

q

Given that

(G,\circ)

is a finite group with

|G|=p^n

, where

p

is prime, and

q

is a prime number different from

p

. To prove that

G

has no subgroup of order divisible by

q

. If possible, let

H

be a subgroup of

G

such that

q

divides

|H|

. By Lagrange's theorem,

\begin{equation*} |H|\mid |G|. \end{equation*}

Therefore

\begin{equation*} |H|\mid p^n. \end{equation*}

Since

q

divides

|H|

, it follows that

q

divides

p^n

. This contradicts that

p

and

q

are distinct prime numbers. Hence,

G

has no subgroup of order divisible by

q

\square

Let

(G,\circ)

be a finite group with

\begin{equation*} |G|=32=2^5. \end{equation*}

Every subgroup of

G

must have order

\begin{equation*} 2^m \end{equation*}

where

0\leq m\leq 5

. Therefore the possible subgroup orders allowed by Lagrange's theorem are

\begin{equation*} 1,2,4,8,16,32. \end{equation*}

A subgroup of order

6

is impossible because

6\nmid 32

Let

(G,\circ)

be a finite group with

\begin{equation*} |G|=27=3^3. \end{equation*}

Every subgroup order must be one of

\begin{equation*} 1,3,9,27. \end{equation*}

Thus no subgroup can have order

6

12

, or

18

Let

(G,\circ)

be a group with

|G|=81

. Prove that

G

has no subgroup of order

12

Let

(G,\circ)

be a group with

\begin{equation*} |G|=81=3^4. \end{equation*}

H

is a subgroup of

G

, then by Lagrange's theorem,

\begin{equation*} |H|\mid 81. \end{equation*}

The positive divisors of

81

are

\begin{equation*} 1,3,9,27,81. \end{equation*}

Since

12

is not in this list,

12\nmid 81

. Therefore

G

has no subgroup of order

12

We use Lagrange's theorem to test whether a proposed subgroup order is even possible. Enter

p

n

, and a candidate order, and compare the candidate with the divisors of

p^n

. If the candidate is allowed, the conclusion is only that Lagrange's theorem does not rule it out. If it is not allowed, then no subgroup of that order can exist in a group of order

p^n

1. Let

|G|=2^6

. List all possible subgroup orders allowed by Lagrange's theorem. 2. Let

|G|=5^3

. Can

G

have a subgroup of order

25

? 3. Let

|G|=7^2

. Can

G

have a subgroup of order

14

1. The possible subgroup orders are

\begin{equation*} 1,2,4,8,16,32,64. \end{equation*}

2. Yes, Lagrange's theorem allows it because

\begin{equation*} 25\mid 125. \end{equation*}

This does not prove that such a subgroup must exist. 3. No. Since

\begin{equation*} 14\nmid 49, \end{equation*}

Lagrange's theorem rules out a subgroup of order

14

They are among

1,p,p^2,\ldots,p^n

No. It only says that other orders are impossible.

Any subgroup order must divide

p^n

, and no prime other than

p

divides

p^n

BMLabs Mathematics

BMLabs Mathematics

Subgroups of Prime Power Order Groups

Subgroups of Prime Power Order Groups

OVERVIEW

PRIME POWER ORDER GROUP

EXPLORE THE PRIME POWER LADDER

PRIME POWER DIVISOR LADDER

HOW THE LADDER USES LAGRANGE

POSSIBLE SUBGROUP ORDERS IN A GROUP OF ORDER THIRTY TWO

POSSIBLE SUBGROUP ORDERS IN A GROUP OF ORDER TWENTY SEVEN

RULING OUT A SUBGROUP IN A PRIME POWER ORDER GROUP

SOLUTION

TEST A PROPOSED SUBGROUP ORDER

PRIME POWER SUBGROUP ORDER TESTER

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation