2.3: Subgroups

Subgroups are non-empty subsets of a group that are themselves groups. In this section, we will define and characterize subgroups.

Definition: Subgroup

Let \(G\) be a group. A non-empty subset \(H\) of \(G\), \(H(\ne \emptyset) \subset G\), is a subgroup of \(G\) if \(H\) is itself a group under the operation inherited from \(G\). We write  \(H \le G\).

 Trivial subgroups of \(G\) are \(\{e\}\) and \(G\) (the group itself).

 

Example \(\PageIndex{1}\)

1.  The trivial subgroups of a group \(G\) are \(\{e\}\) and  \(G.\)
2. \((\mathbb{Z},+)\) is a subgroup of \((\mathbb{R},+).\) In fact \(\mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}\) are subgroups of \(\mathbb{R}\) under addition.
3. But \((\mathbb{Q}^*,\cdot)\) is  not a subgroup of \((\mathbb{R},+).\)  (why?)

Caution

A subgroup is a smaller group that uses exactly the same operation as its parent group. For example, even integers form a group under addition, but positive integers do not because they lack additive inverses.

Note

To ensure a set is a group, we need to check for the following five conditions:

1. Non-empty set,

2. Closure, 

3. Associativity,

4. Identity, and

5. inverse.

The following theorem allows us to check three conditions (rather than five) to determine whether a subset is a subgroup.

Theorem \(\PageIndex{1}\)

 

Let \((G, \star )\) be a group then \(H \subseteq G\) is a subgroup if and only if

The following conditions are satisfied:

1. Identity Containment: The identity \(e \in G\) is in \(H\). ( i.e. \(H\) is a non empty subset of \(G\))

2. Closure under Operation: If \(h_1,h_2 \in H\), then \(h_1 \star h_2 \in H\).

3. Closure under Inverses: If \(h \in H\) then \(h^{-1} \in H\).

 

Proof

Let \((G, \star)\) be a group. 

 Assume that \(H\) is a subgroup of \(G, (H \le G)\). By the definition of a subgroup, \((H, \star)\) is itself a group under the binary operation restricted from \(G\).  Therefore:

1. \(H\) contains an identity element \(e_H \in H\) satisfying \(e_H \star h = h\) for all \(h \in H\). Since \(e_H \in G\), we have \(e_H \star e_H = e_H = e_H \star e_G\). Applying the left cancellation law in \(G\), we conclude \(e_H = e_G\). Thus, \(e_G \in H\).

2. \(H\) is closed under \(\star\) by the definition of a binary operation on \(H\).

3. Let \(h\in H\). Then \(h^{-1} \in G\) and it is unique. Since \(H \le G\), \(h^{-1} \in H\).

 

\(\Leftarrow\)   Conversely, assume that \(H \subseteq G\) satisfies the three conditions. Then

Closure: By Condition 2, \(H\) is closed under \(\star\).

Associativity: Since the operation \(\star\) is associative on \(G\), and \(H \subseteq G\)), associativity holds for all elements in \(H \) automatically (it is inherited from \(G \).
Identity Element: By Condition 1, \(e_G \in H\).  For any \(h \in H , e_G \star h = h \star e_G = h \), so \(e_G\) serves as the identity element of \(H\).
Inverses: By Condition 3, every element \(h \in H\)  has its inverse \(h^{-1} \in H\).
  Since \(H\) is non-empty and satisfies all four group axioms, \(H\) is a group under \(\star\). Therefore, \(H \le G\). \(\square\)

The following theorem allows us to check two conditions (rather than 3) to ensure a subset is a subgroup.

Theorem \(\PageIndex{2}\)

 Let \((G, \star)\) be a group.

Then \(H (\subseteq G)\) is a subgroup of \(G\) iff the following conditions are satisfied:

1.  \(e_G \in H\) ( i.e. \(H\) is a non empty subset of \(G\)), and

2.  \(g \star h^{-1} \in H, \forall \; g,h \in H\).

Proof:

Let \((G, \star)\) be a group. 

 Assume that \(H\) is a subgroup of \(G, (H \le G)\). By the definition of a subgroup, \((H, \star)\) is itself a group under the binary operation restricted from \(G\).  Therefore:

1. \(H\) contains an identity element \(e_H \in H\) satisfying \(e_H \star h = h\) for all \(h \in H\). Since \(e_H \in G\), we have \(e_H \star e_H = e_H = e_H \star e_G\). Applying the left cancellation law in \(G\), we conclude \(e_H = e_G\). Thus, \(e_G \in H\).

2.  Let \(g, h\in H\). Then \(h^{-1} \in G\) and it is unique. Since \(H \le G\), \(h^{-1} \in H\).  Since the binary operation \(\star\) is closed in\(H\), \(g \star h^{-1} \in H\).

 

\(\Leftarrow\)   Conversely, assume that \(H \subseteq G\) satisfies the two conditions. Then

Identity Element: By Condition 1, \(e_G \in H\).  For any \(h \in H , e_G \star h = h \star e_G = h \), so \(e_G\) serves as the identity element of \(H\).

Closure: By Condition 2, \(H\) is closed under \(\star\).

Closure under Inverses: Let \(h \in H\). Choosing \(g = e_G \in H\), Condition 2 guarantees:
    \(  e_G \star h^{-1} = h^{-1} \in H. \)  Thus, every element in \(H\) has its inverse in \(H\).
   Closure under Operation: Let \(g, h \in H\). Since \(h^{-1} \in H\), we apply Condition 2 to \(g\) and \(h^{-1}\),
    \( g \star (h^{-1})^{-1} = g \star h \in H. \) Thus, \(H\) is closed under the group operation \(\star\).
    Associativity: Since the operation \(\star\) is associative on \(G\), and \(H \subseteq G\)), associativity holds for all elements in \(H \) automatically (it is inherited from \(G \).

  Since \(H\) is non-empty and satisfies all four group axioms, \(H\) is a group under \(\star\). Therefore, \(H \le G\). \(\square\)

 

 

Example \(\PageIndex{2}\)

Consider \((\mathbb{R}^*, \bullet)\).  Decide if the following sets form subgroups of \(\mathbb{R}^*\).

1. \(H=\{x\in \mathbb{R}^*|x=1\) or \( x\) is irrational \(\}\).

No.

Counterexample:

Consider \(\sqrt{2} \in H\).

Further \(\sqrt{2} \bullet \sqrt{2}=2\) but \(2 \not \in H\) 

Therefore since there is no closure, \(H \not \le \mathbb{R}^*\). \(\square\)

 

2. \(K = \{x \in \mathbb{R}^*| x \ge 1\}\).

No.

Counterexample:

Consider \(2 \in K\).

Note that \(2^{-1}=\frac{1}{2}\), but \(\frac{1}{2} \not \in K\).

Thus,, the inverse of \(2\) does not exist in \ (K \).

Since \(a^{-1} \not \exists \forall a \in K\), \(K \not \le \mathbb{R}^*\).

 

Example \(\PageIndex{3}\)

 Let \(G = \Big( M_{22}(\mathbb{R), + \Big) }\).

Suppose you take a matrix whose trace is zero.

Thus \(H=\Big\{ \begin{bmatrix} a & b\\
c & d \end{bmatrix} \Big| a+d=0 \Big\}\).

Show that \(H \le G\).

Solution:

We show that \(H\) satisfies the additive subgroup test: the zero matrix is in \(H\), and the difference of any two trace-zero matrices also has trace zero.

Note that \(\begin{bmatrix} 0 & 0\\
0 & 0 \end{bmatrix}\) is the identity of \(G\) and it is clearly an element of \(H\).  Thus, condition (1) of Theorem 2 is satisfied.

  We shall show that \(g \star h^{-1} \in H, \forall \; g,h \in H \). Note that the binary operation is \(+\).

Let \(\begin{bmatrix} a & b\\
c & d \end{bmatrix}, \begin{bmatrix} e & f\\
g & h \end{bmatrix} \in H\).

Then \(a+d=0\) and \( e+h=0\) since the two matrices belong to \(H\).

Consider  \(\begin{bmatrix} a & b\\
c & d \end{bmatrix} + \begin{bmatrix} -e & -f\\
-g & -h \end{bmatrix}=\begin{bmatrix} a-e & b-f\\
c-g & d-h \end{bmatrix} \).

Then \( (a-e)+(d-h)=(a+d)-(e+h)=0-0=0\).

Hence \(\begin{bmatrix} a-e & b-f\\
c-g & d-h \end{bmatrix} \in H\).

Thus \(H \le G\).

Example \(\PageIndex{4}\)

Let \( GL_2 (\mathbb{R})= \Big\{ A=\begin{bmatrix}a & b\\
c & d\end{bmatrix} \big| a,b,c,d \in \mathbb{R}, \det {(A)} =ad-bc \ne 0 \Big\} \) where \(GL_2(\mathbb{R})\) stands for General Linear Group of \(2 \times 2\) invertible matrices  over \(\mathbb{R}\) and  \(SL_2(\mathbb{R})= \Big\{ A=\begin{bmatrix}a & b\\
c & d\end{bmatrix} \big| a,b,c,d \in \mathbb{R}, \det {(A)} =ad-bc = 1 \Big\}\), where \(SL_2(\mathbb{R})\) stands for special Linear Group of \(2 \times 2\) matrices with determinant \(1\) over \(\mathbb{R}\) . Show that \(SL_2( \mathbb{R}) \le GL_2(\mathbb{R})\). 

Answer

Non-emptiness and Subset Property: Consider the \(2 \times 2\) identity matrix:
    \(
    I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
    \)
    The entries are real numbers and \(\det(I) = (1)(1) - (0)(0) = 1\). Therefore, \(I \in SL_2( \mathbb{R})\). Thus \(SL_2( \mathbb{R}) \neq \emptyset.\)

Closure Under Multiplication:
    Let \(A, B \in SL_2(\mathbb{R})\). By definition, \(\det(A) = 1\) and \(\det(B) = 1\). Using the multiplicative property of determinants:
    \(
    \det(AB) = (\det(A))(\det(B) )= (1) ( 1 )= 1.
    \)
    Since \(AB\) is a \(2 \times 2\) matrix with real entries with \(\det(AB) = 1, AB \in SL_2(\mathbb{R})\).

   Closure Under Inverses:
    Let \(A \in SL_2(\mathbb{R})\). Then \(\det(A) = 1 \neq 0\), so \(A^{-1}\) exists in \(GL_2(\mathbb{R})\). Using the inverse property of determinants:
    \(
    \det(A^{-1}) = \dfrac{1}{\det(A)} = \frac{1}{1} = 1.
    \)
    Since \(A^{-1}\) is a \(2 \times 2\) matrix with real entries and \(\det(A^{-1}) = 1\), we have \(A^{-1} \in SL_2(\mathbb{R})\).
 

By the Three-Steps Subgroup Test, \(SL_2(\mathbb{R})\) is a subgroup of \(GL_2(\mathbb{R})\), denoted \(SL_2(\mathbb{R}) \le GL_2(\mathbb{R})\).

 

Example \(\PageIndex{5}\)

Let \(n\) be a nonnegative integer. Then  show that \(n\mathbb{Z}=\{nm|m\in \mathbb{Z}\} \) is a subgroup of \((\mathbb{Z},+).\)

Answer

Let \(n\) be a nonnegative integer. Notice that the identity \(0=(n)(0)\in n\mathbb{Z}\).

Let \(x,y \in n\mathbb{Z}\). Then \(x=na\) and \(y= nb\) for some \(a,b \in n\mathbb{Z}\). Consider \(x-y=na-nb=n(a-b) \). Since \(a-b\in \mathbb{Z}\),  \(x-y \in n\mathbb{Z}\). Therefore\( n\mathbb{Z}\) is a subgroup of \((\mathbb{Z},+).\)

Theorem \(\PageIndex{3}\): Subgroups of integers

Every subgroup \(H\) of \((\mathbb Z,+)\) has the form \(n\mathbb Z\) for a unique integer \(n\ge0\).

Proof

Let \(H\)  be a subgroup of \((\mathbb Z,+)\) .

If \(H=\{0\}\), take \(n=0\). Otherwise \(H\) contains a nonzero integer, so it contains a positive integer by closure under additive inverses. Let \(n\) be its least positive member. Every multiple of \(n\) lies in \(H\).

Conversely,  for \(h\in H\) write \(h=qn+r\), \(0\le r<n\). Since \(h,qn\in H\), \(r=h-qn\in H\). Minimality of \(n\) forces \(r=0\). Therefore \(h\in n\mathbb Z\), giving \(H=n\mathbb Z\). Uniqueness follows from \(n\) being the least positive member when \(n>0\). \(\square\)

 

Example \(\PageIndex{6}\)

Show \(5\mathbb Z\cap 8\mathbb Z=40\mathbb Z\).
 (An integer is in the intersection exactly when divisible by both \(5\) and \(8\), equivalently by their least common multiple \(40\).)

Answer

Let \(x \in 5\mathbb{Z} \cap 8\mathbb{Z}\).
Then \(x = 5a\) and \(x = 8b\) for some
\(a,b \in \mathbb{Z}\).

Thus, \(5a = 8b\). Since \(5\) and \(8\) are
relatively prime, \(8 \mid a\). Hence,
\(a = 8k\) for some \(k \in \mathbb{Z}\).

Therefore,
\(
x = 5a = 5(8k) = 40k.
\)
Thus, \(x \in 40\mathbb{Z}\), and so
\(
5\mathbb{Z} \cap 8\mathbb{Z}
\subseteq 40\mathbb{Z}.
\)

Conversely, let \(x \in 40\mathbb{Z}\).
Then \(x = 40k\) for some \(k \in \mathbb{Z}\).
We have
\(
x = 5(8k) = 8(5k).
\)
Since \(5k,8k \in \mathbb{Z}\), it follows that
\(x \in 5\mathbb{Z}\) and \(x \in 8\mathbb{Z}\).
Hence,
\(
40\mathbb{Z} \subseteq 5\mathbb{Z} \cap 8\mathbb{Z}.
\)

Therefore,
\(
\boxed{5\mathbb{Z} \cap 8\mathbb{Z} = 40\mathbb{Z}}.
\)
\(\square\)

 The finite subgroup test

Definition: finite group

Let \(G\) be a group.  If \(G\) is finite and \(|G|=n\), then we say that the order of the group \(G\) is \(n\), and denoted by \(|G|=n\).

 Let \(G\) be a group. For a finite nonempty subset of \(G\), closure alone can force the other properties.

Theorem \(\PageIndex{3}\)

Subgroup test for finite groups:

Let \(G\) be a finite group, and let \(H\) be a subset of \(G\).

Then \(H \le G\) iff

1. \(H \ne \{\}\). (non-empty).

2. \(a, b \in H\) then \(ab \in H\) (closed).

Proof

Let \(G\) be a finite group, and let \(H\) be a subset of \(G\). 

Assume that \(H \le G\). Then   \(H \ne \{\}\). (non-empty) and \(a, b \in H\) then \(ab \in H\) (closed).

Conversely, suppose \(H\) is finite, nonempty, and closed under multiplication.

Choose \(h\in H\). Then \(h,h^2,h^3,\dots\) all belong to \(H\).  Since \(H\) is finite, two powers must repeat: \(h^r=h^s\) for some integers \(r>s\ge1\). Cancel \(h^s\) in \(G\) to get
\(
h^{r-s}=e.
\)
Since \(r-s\ge1\), this shows \(e\in H\). Moreover,
\(
h^{-1}=h^{r-s-1}\in H,
\)
where if \(r-s=1\), the right-hand side is \(h^0=e\in H\). Thus \(H\) has the identity and inverses, and the three-condition test applies. \(\square\)

Example \(\PageIndex{6}\)

Find all subgroups of  Klein 4-group \( (K_4) \). The  Group \(K_4=\{e, a, b, c\}\) is such that \(a^2=b^2=c^2=e\), with the product of any two distinct non-identity elements is the third: \(ab = c\), \(ac = b\), and \(bc = a\). 

Solution

Since \(K_4\) is a finite Group, \(H_1=\{e,a\}, H_2=\{e,b\}\) and \(H_3=\{e,c\}\) are subgroups of \(K_4.\) Any subgroup of \(K_4\) with more than \(3\) elements must contain two distinct non-identity elements. Therefore, it must contain the third. Hence \(\{e\}, H_1=\{e,a\}, H_2=\{e,b\}\) and \(H_3=\{e,c\}, K_4\) are subgroups of \(K_4.\)

Caution

In an infinite group, nonempty + closure is not enough. The set \(\{0,1,2,\ldots\}\) is closed under addition in \(\mathbb Z\), but \(-1\) is missing.

Overview of Subgroup Tests

Let \(G\) be a group and \(H\) be a subset of \(G \). Then 

Test Name	Prerequisites	Condition(s) to Verify
Two-Steps Test	\(H \neq \emptyset, H \subseteq G\)	\(\forall a, b \in H, \; a b^{-1} \in H\)
Three-Steps Test	\(H \neq \emptyset, H \subseteq G\)	\(\forall a, b \in H, a b \in H\) \(\forall a \in H, a^{-1} \in H\)
Finite Subgroup Test	\(H \neq \emptyset, H \subseteq G, |G| < \infty\)	\(\forall a, b \in H, \; a b \in H\) (Closure only)

Lattice Diagram of all subgroups

A Lattice Diagram of Subgroups (often called a subgroup lattice) is a visual graph that shows all the subgroups of a given group and how they relate to one another through inclusion (the subgroup relation \(\le\)).

Key Rules of a Subgroup Lattice:

Nodes (Vertices): Each node represents a distinct subgroup of \(G\).

Levels (Hierarchy): Subgroups with larger cardinality (order) are placed higher on the page. The full group \(G\) is always at the very top. The trivial subgroup \(\{e \} \) is always at the very bottom.

Edges (Connections): A line is drawn connecting a lower subgroup \(H\) to a higher subgroup \(K\) (\(H \le K\)) if \(H\) is a maximal subgroup of \(K\) (meaning there is no intermediate subgroup \(M\) strictly between them, \(H \subsetneq M \subsetneq K\)).

Transitivity: If there is a path from \(A\) to \(B\) and from \(B\) to \(C\), then \(A \le C\), even if there is no direct line between \(A\) and \(C \).

Example \(\PageIndex{7}\): Lattice diagram of subroups for \(\mathbb{Z}_6\) 

Example \(\PageIndex{8}\): Lattice diagram of subgroups for Klein four group \(K_4\)

 

Special subgroups 

Some important subgroups collect elements based on how they interact with others. For these definitions, we write the group operation as \(ab\), instead of \(a \star b\).
 

Centralizer of an element

Definition: Centralizer of an element

Let \(G\) be a group.  For  \(a \in G\), the Centralizer of \(a\) in the group \(G\),  denote by  \(C_G(a)\) or \(C(a)\), \(C(a)=\{g \in G|ag=ga\}\). 

 

Theorem \(\PageIndex{4}\)

Let \(G\) be a group.  For any \(a \in G\), the Centralizer of \(a\) in the group \(G\),  \(C(a)=\{g \in G|ag=ga\}\). Then  \(C(a)\le G\).

Proof

If \(C(a)=\{g \in G|ag=ga\}=G\), then we are done. Otherwise, let \(x_1, x_2 \in C(a) \) .

We will show that \(x_1x_2^{-1} \in C(a) \) and \(e_G \in C(a) \) .

Consider \begin{align*} ax_1 &=x_1a\\ ax_1x_2^{-1} &=x_1ax_2^{-1} \\ ax_1x_2^{-1} &=x_1(ax_2^{-1}) \\ ax_1x_2^{-1}&=x_1(x_2^{-1}a) \\ax_1x_2^{-1} &=x_1x_2^{-1}a \end{align*} , \(x_1,x_2^{-1} \in C(a) \) .

Since \(x_2,x_2^{-1} \in C(a) \) , \(x_2x_2^{-1}=e \) , where \(e \) is the identity element, \(e \) in \(C(a) \) .

Since \(x_1x_2^{-1}, e \in C(a) \) , \(C(a) \le G \) . \(\square\)

Example \(\PageIndex{9}\)

Let \(G=GL(2,\mathbb{R})\).  Then find

1. \(C \Bigg( \begin{bmatrix}1 & 1\\
   1 & 0 \end{bmatrix}\Bigg) \).

2. \(C \Bigg( \begin{bmatrix} 0 & 1\\
   1 & 0 \end{bmatrix}\Bigg)\).

Solution:

Let \(\begin{bmatrix} a & b\\
c & d \end{bmatrix} \in G\), Then \(a,b,c,d \in \mathbb{R} \), and \(ad-bc \ne 0.\)

Consider \(C \Bigg( \begin{bmatrix}1 & 1\\
1 & 0 \end{bmatrix}\Bigg)\) means that \(\begin{bmatrix}1 & 1\\
1 & 0 \end{bmatrix}\begin{bmatrix}a & b\\
c & d \end{bmatrix}=\begin{bmatrix}a & b\\
c & d \end{bmatrix}\begin{bmatrix}1 & 1\\
1 & 0 \end{bmatrix}\).

Thus \(\begin{bmatrix}a+c & b+d\\
a & b \end{bmatrix}=\begin{bmatrix}a+b & a\\
c+d & c \end{bmatrix} \).

Thus \(a+b=a+c\), \(a=b+d\), \(c+d=a\) and \(c=b\).

Thus \(C\Bigg(\begin{bmatrix} 1 & 1\\
1 & 0 \end{bmatrix}\Bigg) =\Bigg\{A= \begin{bmatrix} a & b\\
b & a-b \end{bmatrix} \Bigg| a,b \in \mathbb{R}, \det(A) \ne 0 \; \Bigg\}\). 

2. \(C \Bigg( \begin{bmatrix} 0 & 1\\
   1 & 0 \end{bmatrix}\Bigg)\).

Let \(\begin{bmatrix} a & b\\
c & d \end{bmatrix} \in G\), Then \(a,b,c,d \in \mathbb{R} \), and \(ad-bc \ne 0.\)

Consider \(C \Bigg( \begin{bmatrix}0 & 1\\
1 & 0 \end{bmatrix}\Bigg)\) means that \(\begin{bmatrix}0 & 1\\
1 & 0 \end{bmatrix} \begin{bmatrix}a & b\\
c & d \end{bmatrix}=\begin{bmatrix}a & b\\
c & d \end{bmatrix}\begin{bmatrix}0 & 1\\
1 & 0 \end{bmatrix}\).

Thus \(\begin{bmatrix}c & d\\
a & b \end{bmatrix}=\begin{bmatrix}b & a\\
d & c \end{bmatrix} \).

Thus \(b=c\) and \(d=a\).

Thus \(C\Bigg(\begin{bmatrix} 0 & 1\\
1 & 0 \end{bmatrix}\Bigg) =\Bigg\{B= \begin{bmatrix} a & b\\
b & a \end{bmatrix} \Bigg| a,b \in \mathbb{R}, \det(B) \ne 0 \; \Bigg\}\). 

Center of a group

Definition: Center

Let \(G\) be a group. Then the center of \(G\) is defined as \(Z(G)=\{x \in G \mid gx=xg \forall \; g \in G \}.\) The center is the set of elements that commute with every element. 

 

Theorem \(\PageIndex{4}\)

 

Let \((G, \star )\) be a group then \(Z(G)\) is a subgroup of \(G.\)

Proof:

Let \(G\) be a group and \(g \in G\).

We will show that \(Z(G)=\{x \in G| gx=xg \; \forall g \in G \}\).

 

Let \(e\) be the identity of \(G\).

We will show that \(e \in Z(G)\).

Since \(eg=ge \; \forall g \in G\), \(e \in Z(G)\).

 

Let \(x_1, x_2 \in Z(G)\).

We shall show that \(x_1x_2^{-1} \in Z(G)\).

Consider that \(x_2^{-1}g=gx_2^{-1}\).

Further \((x_1x_2^{-1})g=x_1gx_2^{-1}\).

     \(=(x_1g)x_2^{-1}\)

     \(=(gx_1)x_2^{-1}\)

     \(=g(x_1x_2^{-1})\). \(\square\)

 

Note

Note that if \(G\) is abelian, \(Z(G)=G\) . 

 

Example \(\PageIndex{10}\)

Let \(G=GL(2,\mathbb{R})\).  Then find \({Z}(G)\).

 Solution:

\({Z}(G)= \{ g \in G| gx=xg, \forall x \in G\}\).

Let \(\begin{bmatrix} a & b
c & d \end{bmatrix} \in G\), Then \(a,b,c,d \in \mathbb{R} \), and \(ad-bc \ne 0.\)

We have shown \(C\Bigg(\begin{bmatrix} 1 & 1\\
1 & 0 \end{bmatrix}\Bigg) =\Bigg\{ \begin{bmatrix} a & b\\
b & a-b \end{bmatrix} \Bigg| a,b \in \mathbb{R}, \det \ne 0 \; \Bigg\}\) and \(C\Bigg(\begin{bmatrix} 0 & 1\\
1 & 0 \end{bmatrix}\Bigg) =\Bigg\{ \begin{bmatrix} a & b
b & a \end{bmatrix} \Bigg| a,b \in \mathbb{R}, \det \ne 0 \; \Bigg\}\), \(\begin{bmatrix}1 & 1\\
1 & 0 \end{bmatrix}, \begin{bmatrix}0 & 1\\
1 & 0 \end{bmatrix} \in G\).

Thus for \(\begin{bmatrix} a & b\\
c & d \end{bmatrix}\) to be in \(\mathbb{Z}(G)\), \( \begin{bmatrix} a & b\\
b & a-b \end{bmatrix} = \begin{bmatrix} a & b\\
b & a \end{bmatrix} \).

Thus since \(a-b=a\), \(b=0\).

Thus, \({Z}(G) =\Bigg\{ \begin{bmatrix} a & 0\\
0 & a \end{bmatrix} \Bigg| a \in \mathbb{R}, a \ne 0 \; \Bigg\}\).

Example \(\PageIndex{11}\)

Prove that for each element \(a \in G\), where \(G\) is a group, that \(C(a)=C(a^{-1})\).

Proof:

Let \(x_1 \in C(a)\) s.t. \(ax_1=x_1a\).

We will show that \(x_1a^{-1}=a^{-1}x_1\).

Consider that \(a^{-1}ax_1=a^{-1}x_1a\)

     \(ex_1=a^{-1}x_1a\)

       \(x_1=a^{-1}x_1a\)

\(x_1a^{-1}=a^{-1}x_1aa^{-1}\)

\(x_1a^{-1}= a^{-1}x_1e\)

\(x_1a^{-1}= a^{-1}x_1\).

Thus, \(x_1 \in C(a^{-1})\).

Hence \(C(a^{-1}) \subseteq C(a)\).

Let \(x_2 \in C(a^{-1})\) s.t. \(a^{-1}x_2=x_2a^{_1}\).

Consider that \(aa^{-1}x_2=ax_2a^{-1}\)

     \(ex_2=ax_2a^{-1}\)

       \(x_2=ax_2a^{-1}\)

\(x_2a=ax_2a^{-1}a\)

\(x_2a= ax_2e\)

\(x_2a= ax_2\).

Thus, \(x_2 \in C(a)\).

Hence \(C(a) \subseteq C(a^{-1})\).
Since \(C(a^{-1}) \subseteq C(a)\) and \(C(a) \subseteq C(a^{-1})\), \(C(a) = C(a^{-1})\). \(\square\)

Centralizer

Definition: Centralizer

Let \(G\) be a group and \(H\) be a subgroup of \(G\) then  the Centralizer of \(H\) in \(G\) is  \(C(H)=\{g \in G|gh=hg \; \forall \;h \in H\}\).

 Note that if \(H=\{e\}\) then \(C(H)=G\).

Theorem \(\PageIndex{5}\)

Let \(G\) be a group and \(H\) be a subgroup of \(G\) then  the Centralizer of \(H\) in \(G\) is  \(C(H)=\{g \in G|gh=hg \; \forall \;h \in H\}\) is a subgroup of \(G.\)

Proof:

Let \(G\) be a group and \(H\) be a subgroup of \(G\). We will show that \(C(H)\) is a subgroup of \(G\).

Let \(e\) be the identity element of \(G\), then \(e \in C(H)\).

Let \(g_1,g_2 \in C(H)\).

We will show that \(g_1g_2^{-1}h=hg_1g_2^{-1}\).

Consider that \( g_1h=hg_1\), \(g_2h=hg_2\), \(\forall h \in H\).

Since \(g_1h=hg_1\),

    \(g_1hg_2^{-1}=hg_1g_2^{-1}\)

    \(g_1g_2^{-1}h=hg_1g_2^{-1}\).

Thus \(g_1g_2^{-1} \in C(H)\)

Hence \(C(H) \le G\).

Conjugacy classes of a group \(G\) 

Two elements \(x,y \) in a group \(G\) are conjugate if \(y=gxg^{-1}\) for some \(g\in G\). This is an equivalence relation on \(G\).

Conjugacy classes of \(G\) are subsets of \(G\) whose elements are conjugate to one another. That is, \([x]=\{gxg^{-1}| g\in G \}\), \(x \in G\).

Normalizer

Let \(G\) be a group and \(H\) be a subgroup of \(G\).  For any \(x \in G\), \(N(H)=xHx^{-1}=\{xhx^{-1}|h \in H\}\).

This subgroup is called the Normalizer of \(H\) in the group \(G\).

Definition: Normalizer

Let \(G\) be a group and \(H\) be a subgroup of \(G\). Then 

for any \(x \in G, N_x(H)=xHx^{-1}=\{xhx^{-1}|h \in H\}\) Gonjugates of \(G\), and 

the Normalizer of \(H\) in the group \(G\), \(N(H)= \{ xHx^{-1} | x \in G\}.\)

Theorem \(\PageIndex{6}\)

Let \(G\) be a group and \(H\) be a subgroup of \(G\). Then  for each \(g \in G\), \(N_g(H)=gHg^{-1} \) is a subgroup of \(G.\) These subgroups are called conjugates of \(H\) in \(G\).

Proof:

 Let \(g \in G.\) Let \(e\) be the identity of \(H\) then \(geg^{-1} \in gHg^{-1}\).

Consider \(geg^{-1}=(ge)g^{-1}=gg^{-1}=e\).  Thus \(e \in gHg^{-1}\).

 

Let \(a,b \in gHg^{-1}\).

We will show that \(ab^{-1} \in H\).

Consider \(a=gh_1g^{-1}\) and \(b=gh_2g^{-1}\), \(h_1,h_2 \in H\).

Then \(ab^{-1}=(gh_1g^{-1})(gh_2g^{-1})^{-1}\)

     \(=(gh_1g^{-1})((g^{-1})^{-1}h_2^{-1}g^{-1})\)

     \(=gh_1g^{-1}gh_2^{-1}g^{-1}\)

     \(=gh_1h_2^{-1}g^{-1} \in gHg^{-1}\).

Thus since \(h_1, h_2^{-1} \in N_g(H)\) and \(e \in N_g(H)\), \(N_g(H)\) is a subgroup of \(G\). \(\square\)