Conjugate Subgroups and Invariance

The class equation studies how individual elements move under conjugation. The same idea applies to subgroups: a subgroup can be moved to another subgroup by conjugating all its elements at once. Conjugate subgroups have the same internal group structure, although they may sit in different positions inside the larger group. This distinction is important in Sylow theory, where Sylow subgroups are often not equal but are conjugate, and where normality is detected by invariance under conjugation.

Let

G

be a group, let

H

be a subgroup of

G

, and let

a \in G

. The \textbf{conjugate of

H

a

} is the set

\begin{equation*} aHa^{-1}=\{aha^{-1}\mid h \in H\}. \end{equation*}

Let

G

be a group, let

H

be a subgroup of

G

, and let

a \in G

. The \textbf{conjugate of

H

a

} is the set

\begin{equation*} aHa^{-1}=\{aha^{-1}\mid h \in H\}. \end{equation*}

Let

G

be a group, let

H

be a subgroup of

G

, and let

a \in G

. Then

aHa^{-1}

is a subgroup of

G

Given that

G

is a group,

H

is a subgroup of

G

, and

a \in G

. To prove that

aHa^{-1}

is a subgroup of

G

. Since

e \in H

, we have

\begin{equation*} aea^{-1}=e. \end{equation*}

Therefore

e \in aHa^{-1}

, so

aHa^{-1}

is non-empty. Let

ah_1a^{-1},ah_2a^{-1} \in aHa^{-1}

, where

h_1,h_2 \in H

. Then

\begin{align*} (ah_1a^{-1})(ah_2a^{-1})^{-1} &= (ah_1a^{-1})(ah_2^{-1}a^{-1}) \\ &= a(h_1h_2^{-1})a^{-1}. \end{align*}

Since

H

is a subgroup of

G

, we have

h_1h_2^{-1} \in H

. Therefore

a(h_1h_2^{-1})a^{-1} \in aHa^{-1}

. Hence

aHa^{-1}

is a subgroup of

G

\square

Let

G

be a group, let

H

be a subgroup of

G

, and let

a \in G

. Then

H

is isomorphic to

aHa^{-1}

Given that

G

is a group,

H

is a subgroup of

G

, and

a \in G

. To prove that

H

is isomorphic to

aHa^{-1}

. Define

f:H\to aHa^{-1}

\begin{equation*} f(h)=aha^{-1}. \end{equation*}

Let

h_1,h_2 \in H

. Then

\begin{align*} f(h_1h_2) &= a(h_1h_2)a^{-1} \\ &= (ah_1a^{-1})(ah_2a^{-1}) \\ &= f(h_1)f(h_2). \end{align*}

Therefore

f

is a homomorphism. Suppose that

f(h_1)=f(h_2)

. Then

ah_1a^{-1}=ah_2a^{-1}

. Multiplying on the left by

a^{-1}

and on the right by

a

gives

h_1=h_2

. Therefore

f

is one-one. Let

y \in aHa^{-1}

. Then

y=aha^{-1}

for some

h \in H

, and so

y=f(h)

. Therefore

f

is onto. Therefore

f

is an isomorphism. Hence

H \cong aHa^{-1}

\square

Let

G

be a group, let

H

be a subgroup of

G

, and let

a \in G

. The subgroup

H

is called \textbf{invariant under

a

} if

\begin{equation*} aHa^{-1}=H. \end{equation*}

Let

G

be a group and let

H

be a subgroup of

G

. Then

H

is normal in

G

if and only if

H

is invariant under every element of

G

Given that

G

is a group and

H

is a subgroup of

G

. To prove that

H

is normal in

G

if and only if

H

is invariant under every element of

G

. By definition,

H

is normal in

G

if and only if

\begin{equation*} gHg^{-1}=H \end{equation*}

for every

g \in G

. This is exactly the statement that

H

is invariant under every element of

G

. Hence

H

is normal in

G

if and only if

H

is invariant under every element of

G

\square

Let

G

be a group and let

H,K

be subgroups of

G

. The \textbf{conjugates of

H

induced by

K

} are the subgroups

\begin{equation*} kHk^{-1},\quad k \in K. \end{equation*}

Let

G

be a group and let

H,K

be subgroups of

G

. Then every subgroup of the form

kHk^{-1}

, where

k \in K

, is a conjugate subgroup of

H

G

Given that

G

is a group and

H,K

are subgroups of

G

. To prove that every subgroup of the form

kHk^{-1}

, where

k \in K

, is a conjugate subgroup of

H

G

. Let

k \in K

. Since

K

is a subgroup of

G

, we have

k \in G

. By the definition of a conjugate subgroup,

kHk^{-1}

is a conjugate of

H

by the element

k

. Hence every subgroup of the form

kHk^{-1}

, where

k \in K

, is a conjugate subgroup of

H

G

\square

Let

G=S_3

and let

H=\{e,(12)\}

. Conjugating

H

by elements of

S_3

produces the subgroups generated by the three transpositions:

\begin{equation*} \{e,(12)\},\quad \{e,(13)\},\quad \{e,(23)\}. \end{equation*}

These subgroups are not all equal, but they are conjugate in

S_3

. Each has order

2

, and each is isomorphic to the cyclic group of order

2

This preview shows what happens when a whole subgroup is conjugated at once. Choose a subgroup of

S_3

and see the distinct subgroups obtained from it by conjugation. The transposition subgroup moves among three different order two subgroups, while

A_3

stays fixed under every conjugation. Use the output to distinguish being isomorphic from being equal. This prepares the Sylow idea that subgroups may be different but still conjugate.

The orbit of a subgroup under conjugation records all positions that subgroup can occupy inside the ambient group. If the orbit has only one subgroup, then every conjugation returns the same subgroup. That is exactly normality expressed as invariance.

This calculator checks the normality criterion in the finite examples from this page. Select a subgroup of

S_3

. The output lists whether conjugation fixes the subgroup for each type of conjugating element. The result is not just a yes-or-no test; it explains whether the subgroup is invariant under every element, which is the definition of normality used in the theorem.

A conjugate subgroup is always isomorphic to the original subgroup, but normality asks for equality after every conjugation. The calculator highlights this difference by showing which elements fix the subgroup and which elements move it.

Let

H

be a subgroup of a group

G

. Prove that

H

is normal in

G

if and only if

H

is a union of conjugacy classes of elements of

G

Let

H

be a subgroup of a group

G

. Suppose first that

H

is normal in

G

. Let

h \in H

and let

g \in G

. Since

H

is normal in

G

, we have

gHg^{-1}=H

. Therefore

ghg^{-1}\in H

. Hence

Cl_G(h)\subseteq H

for every

h \in H

. Therefore

\begin{equation*} H=\bigcup_{h\in H}Cl_G(h). \end{equation*}

Thus

H

is a union of conjugacy classes of

G

. Conversely, suppose that

H

is a union of conjugacy classes of

G

. Let

h \in H

and

g \in G

. Since

H

is a union of conjugacy classes and

h \in H

, we have

Cl_G(h)\subseteq H

. Therefore

ghg^{-1}\in H

. Hence

gHg^{-1}\subseteq H

for every

g \in G

. Replacing

g

g^{-1}

gives

g^{-1}Hg\subseteq H

. Multiplying this containment on the left by

g

and on the right by

g^{-1}

gives

H\subseteq gHg^{-1}

. Therefore

gHg^{-1}=H

for every

g \in G

. Hence

H

is normal in

G

\square

Let

G

be an abelian group and let

H

be a subgroup of

G

. Prove that every conjugate of

H

is equal to

H

Let

G

be an abelian group and let

H

be a subgroup of

G

. For

g \in G

and

h \in H

, commutativity gives

gh=hg

. Therefore

\begin{align*} ghg^{-1} &= hgg^{-1} \\ &= h. \end{align*}

Thus

gHg^{-1}\subseteq H

for every

g \in G

. For every

h \in H

, we also have

h=ghg^{-1}

because

ghg^{-1}=h

. Therefore

H\subseteq gHg^{-1}

. Hence

gHg^{-1}=H

for every

g \in G

. Thus every conjugate of

H

is equal to

H

\square

[1] Let

H

be a subgroup of

G

and let

a \in G

. Prove that

|aHa^{-1}|=|H|

when

H

is finite. [2] Let

H

be a normal subgroup of

G

. Find all conjugates of

H

. [3] In

S_3

, let

H=\{e,(123),(132)\}

. Prove that

H

is invariant under every element of

S_3

[1] The map

h \mapsto aha^{-1}

is a bijection from

H

onto

aHa^{-1}

, so the two finite sets have equal cardinality. [2] If

H

is normal in

G

, then

gHg^{-1}=H

for every

g \in G

. Thus the only conjugate of

H

H

itself. [3] The subgroup

H=A_3

has index

2

S_3

, so it is normal. Therefore it is invariant under every element of

S_3

No. Conjugate subgroups are isomorphic, but they need not be equal.

It means conjugating the subgroup by that element returns the same subgroup.

A subgroup is normal exactly when it is invariant under every element of the ambient group.

BMLabs Mathematics

BMLabs Mathematics

Conjugate Subgroups and Invariance

Conjugate Subgroups and Invariance

CONJUGATE SUBGROUP

INVARIANCE UNDER AN ELEMENT

CONJUGATES INDUCED BY A SUBGROUP

EXPLORE CONJUGATE SUBGROUPS IN S3

CONJUGATE SUBGROUP ORBIT EXPLORER

INTERPRET THE ORBIT

TEST INVARIANCE UNDER CONJUGATION

INVARIANCE TEST CALCULATOR

CONNECT THE TEST

FREQUENTLY ASKED QUESTIONS

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Coordinates and Locus

Invariants under Orthogonal Transformation