Subgroups of Finite p-Groups

The structure theory of finite p-groups does not stop with the nontrivial center. The focus keyword is subgroups of finite p-groups, because finite p-groups contain subgroups of every possible p-power order below their own order. This is a strong internal existence theorem and a conceptual bridge from Cauchy's theorem to Sylow theory. Before finding large p-subgroups inside arbitrary finite groups, we first learn how p-subgroups grow inside p-groups.

Let

G

be a finite p-group of order

p^n

. A \textbf{subgroup chain in a p-group} is a sequence

\begin{equation*} \{e\}=H_0\leq H_1\leq\cdots\leq H_n=G \end{equation*}

such that

|H_i|=p^i

for every

i

satisfying

0\leq i\leq n

Let

G

be a finite p-group of order

p^n

. A \textbf{subgroup chain in a p-group} is a sequence

\begin{equation*} \{e\}=H_0\leq H_1\leq\cdots\leq H_n=G \end{equation*}

such that

|H_i|=p^i

for every

i

satisfying

0\leq i\leq n

Let

G

be a nontrivial finite p-group. Then

G

has a normal subgroup of order

p

Given that

G

is a nontrivial finite p-group. To prove that

G

has a normal subgroup of order

p

. By the nontrivial center theorem,

Z(G)

is nontrivial. Since

Z(G)

is a finite p-group, Cauchy's theorem gives an element

z\in Z(G)

such that

o(z)=p

. Let

N=\langle z\rangle

. Then

|N|=p

. Since

z\in Z(G)

, every element of

N

commutes with every element of

G

. Therefore

N\leq Z(G)

. Every subgroup of the center is normal in

G

. Hence

G

has a normal subgroup of order

p

\square

Let

G

be a finite p-group of order

p^n

. Then for every integer

i

satisfying

0\leq i\leq n

, the group

G

has a subgroup of order

p^i

Given that

G

is a finite p-group of order

p^n

. To prove that for every integer

i

with

0\leq i\leq n

, the group

G

has a subgroup of order

p^i

. We use induction on

n

. If

n=0

, then

G=\{e\}

, and the only required subgroup has order

1=p^0

. Assume the result holds for all finite p-groups of order smaller than

p^n

, where

n\geq1

. By the lemma,

G

has a normal subgroup

N

of order

p

. Then

G/N

is a finite p-group of order

p^{n-1}

. Let

i

satisfy

0\leq i\leq n

. If

i=0

, then

\{e\}

has order

1=p^0

. If

i\geq1

, then

i-1

satisfies

0\leq i-1\leq n-1

. By the induction hypothesis applied to

G/N

, there is a subgroup

K/N\leq G/N

such that

|K/N|=p^{i-1}

. The inverse image

K

is a subgroup of

G

containing

N

, and

\begin{align*} |K|&=|K/N||N|\\ &=p^{i-1}\cdot p\\ &=p^i. \end{align*}

Hence

G

has a subgroup of order

p^i

for every

0\leq i\leq n

\square

Let

G

be a finite p-group and let

H\leq G

. If

[G:H]=p

, then

H

is normal in

G

Given that

G

is a finite p-group and

H\leq G

with

[G:H]=p

. To prove that

H

is normal in

G

. Let

G

act by left multiplication on the set

G/H

of left cosets of

H

. This action gives a homomorphism

\varphi:G\to S_p

. The kernel of this action is contained in

H

, because any element fixing the coset

H

must lie in

H

. The image

\varphi(G)

is a p-subgroup of

S_p

, so

|\varphi(G)|

divides both

|G|

and

p!

. The highest power of

p

dividing

p!

p

, so

|\varphi(G)|

1

p

. The action on

G/H

is transitive, so

p

divides

|\varphi(G)|

. Therefore

|\varphi(G)|=p

. Hence

[G:\ker\varphi]=p

. Since

\ker\varphi\leq H

and

[G:H]=p

, it follows that

\ker\varphi=H

. Since kernels of homomorphisms are normal,

H

is normal in

G

. Hence

H\trianglelefteq G

\square

Let

G=\mathbb{Z}_{27}

under addition modulo

27

. Since

|G|=27=3^3

, the theorem guarantees subgroups of orders

1,3,9,27

. These are

\begin{equation*} \{0\},\quad \langle9\rangle,\quad \langle3\rangle,\quad \mathbb{Z}_{27}. \end{equation*}

Hence

\mathbb{Z}_{27}

has a full subgroup chain.

This preview displays the subgroup orders guaranteed inside a finite p-group of order

p^n

. Choose

p

and

n

, then follow the chain from the trivial subgroup to the whole group. Each step multiplies the order by

p

, matching the induction proof through a normal subgroup of order

p

. Try the presets for

\mathbb Z_{27}

and

\mathbb Z_{16}

, then change

n

to see how the chain grows.

The theorem guarantees existence, not uniqueness. The displayed chain shows the required orders, while the actual subgroups may depend on the group. The proof constructs them by passing through a central normal subgroup and lifting subgroups from a quotient.

This calculator tests a requested subgroup order inside a finite p-group of order

p^n

. Enter

p

n

, and a target exponent

i

. The calculator lists all guaranteed orders and tells you whether

p^i

is included. Use the default to confirm that a group of order

32=2^5

has a subgroup of order

8=2^3

The calculator is a fast way to read the theorem: every exponent from

0

n

occurs. It does not identify the subgroup itself, but it confirms which orders the theorem guarantees.

Let

G

be a group of order

32

. Prove that

G

has a subgroup of order

8

Given that

|G|=32=2^5

. Therefore

G

is a finite

2

-group. By the subgroup theorem for finite p-groups,

G

has a subgroup of order

2^i

for every

0\leq i\leq5

. Taking

i=3

gives a subgroup of order

2^3=8

. Hence

G

has a subgroup of order

8

Let

G

be a finite p-group of order

p^4

. Prove that

G

has a normal subgroup of order

p^3

Given that

|G|=p^4

. By the subgroup theorem for finite p-groups, there exists a subgroup

H\leq G

such that

|H|=p^3

. Then

\begin{align*} [G:H]&=\frac{|G|}{|H|}\\ &=\frac{p^4}{p^3}\\ &=p. \end{align*}

By the theorem that subgroups of index

p

in finite p-groups are normal,

H\trianglelefteq G

. Hence

G

has a normal subgroup of order

p^3

1. Let

|G|=p^5

. Which subgroup orders are guaranteed? 2. Prove that every group of order

16

has a subgroup of order

4

. 3. Let

G

be a finite p-group and

H\leq G

with

|H|=|G|/p

. Prove that

H

is normal. 4. Give the subgroup chain in

\mathbb{Z}_{16}

1. The guaranteed subgroup orders are

1,p,p^2,p^3,p^4,p^5

. 2. Since

16=2^4

, the subgroup theorem gives a subgroup of order

2^2=4

. 3. The condition

|H|=|G|/p

gives

[G:H]=p

, so

H

is normal in

G

. 4. One chain is

\{0\}\leq\langle8\rangle\leq\langle4\rangle\leq\langle2\rangle\leq\mathbb{Z}_{16}

No. Inside an arbitrary finite group, p-subgroups need not be normal. Inside a finite p-group, subgroups of index

p

are normal.

The center gives a normal subgroup of order

p

, which allows induction through a quotient.

No. It proves existence of subgroups of each p-power order, not a complete classification.

BMLabs Mathematics

BMLabs Mathematics

Subgroups of Finite p-Groups

Subgroups of Finite p-Groups

WHY P-GROUPS CONTAIN MANY SUBGROUPS

SUBGROUP CHAIN IN A P-GROUP

BUILD THE GUARANTEED SUBGROUP CHAIN

SUBGROUP CHAIN BUILDER

INTERPRET THE CHAIN

CHECK GUARANTEED P-SUBGROUP ORDERS

P-GROUP SUBGROUP ORDER CALCULATOR

CONNECT THE CALCULATOR

FREQUENTLY ASKED QUESTIONS

CONNECT CAUCHY'S THEOREM TO NORMALITY

Coordinates and Locus

Invariants under Orthogonal Transformation