2.3: Subgroups

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

Definition: Subgroup

Let \(G\) be a group. Let \(H\) is a non empty subset of \(G\) and \((H,\ast)\) is a group, then \(H\) is a subgroup of \(G\).  We write this as \(H \le G\).

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

 

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 5) to ensure a subset is a subgroup.

 

Theorem \(\PageIndex{1}\)

 

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

The following conditions are satisfied:

 

1. The identity \(e \in G\) is in \(H\).

2. If \(h_1,h_2 \in H\), then \(h_1 \ast h_2 \in H\).

3. If \(h \in H\) then \(h^{-1} \in H\).

 

Proof

Assume that \(H \subseteq G\).

We shall show that \(e_H=e_G\), where \(e_H\) is the identity of \(H\) and \(e_H\) is the identity of \(G\).

 

Thus, it is clear that \(H \le G\).◻

•  

  1. From (2), we have shown \(h_1 \ast h_2 \in H, \forall h_1,h_2 \in H\).

  2. From (1), we have shown the identity.

  3. From (3), we have shown the inverses exist.

  4. Every element of \(h_1,h_2,h_3 \in G\) therefore associativity applies because \(H\) inherits it from \(G\) by definition.

•  

  Conversely, assume that \(H \subseteq G\) satisfies 1, 2, and 3 shown in Theorem 1.

  From (1), we have shown that \(H\) is a non-empty set - it has at least an identity.

  Since \(e_G\) is the identity of \(G\) and \(e_H \in G\), \(e_Ge_H=e_He_G=e_H\).

  Further, \(e_H \in H, e_H^{-1}=e_H \in H\).

  Since \(e_H, e_H^{-1} \in G\), \(e_He_H^{-1}=e_G\), but \(e_He_H^{-1}=e_H\).  

  Hence \(e_G=e_H\).◻

   

  Note: all subgroups of a group have a common element - the identity \(e\). They could have more common elements. Since \(H\) is a subset of \(G\), \(e_H\) is in \(H\) by default.

  2. We shall show that, if \(h_1,h_2 \in H\), then \(h_1 \ast h_2 \in H\).

   

  Let \(h_1, h_2 \in H\).

  Since \((H, \ast)\) is a group, \(h_1 \ast h_2 \in H\).◻

  Note a group is closed by definition.

   

  3. We shall show that,if \(h \in H\) then \(h^{-1} \in H\).

   

  Since \((H, \ast)\) is a group, \(h^{-1} \in H\).◻

  Note each element in a group has an inverse by definition.

   

  \(H \le G\) iff:

  \(H \ne \{\}\).

  \((H, \ast)\) is a group.  Need to check:

  1. ,\(h_1 \ast h_2 \in H, \forall h_1,h_2 \in H\)

  2. \(e \in H\), where \(e \in G\),

  3. \(h^{-1} \in H, \forall \; h \in H\),

  4. \(h_1(h_2h_3)=(h_1h_2)h_3, \forall \; a,b,c \in H\).

 

 

The following theorem allows us to check two conditions (rather than 5) to ensure a subset is a subgroup.

Theorem \(\PageIndex{2}\)

 Let \((G, \ast)\) be a group.

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

1.  \(e_G \in H\), and

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

Proof:

Assume that \(H \le G\).

Assume that \(H \subseteq G\).

 

Since \(e \ast h_2 \in H, e \ast h_2^{-1} \in H\).

Further, \(eh_2^{-1}=h_2^{-1} \in H\).

Since \(h_1 \ast h_2 \in H, \ h_1 \ast (h_2^{-1})^{-1} \in H\).

Therefore \(h_1 \ast h_2 \in H\). ◻

It has an identity since \(H \ne \{\}\).

See proof of (a) above

Associativity comes because \(h_1,h_2,h_3 \in H\).

 

1. Let \(h_1, h_2 \in H\).

 

Conversely, assume that \(H \subseteq G\) satisfies (1) and (2) of Theorem 2.

We shall show \(H \le G\).

\(H \ne \{\}\).  Since we have a group, we have at least an identity.

\(h_1 \ast h_2 \in H, \forall h_1,h_2 \in H\),

\(e \in H\), where \(e \in G\),

\(h^{-1} \in H, \forall \; h \in H\),

\(h_1(h_2h_3)=(h_1h_2)h_3, \forall \; a,b,c \in H\).

We shall show that \(g \ast h^{-1} \in H, \forall \; g,h \in H\).

Then \(h^{-1} \in H\) because \(H \subseteq G\).

Since \(g,h^{-1} \in H\), \(gh^{-1} \in H\),

Hence the result.◻

\(H \le G\) iff:

\(H \ne \{\}\).

\((H, \ast)\) is a group.  Need to check:

We shall show that \(e_H=e_G\), where \(e_H\) is the identity of \(H\) and \(e_H\) is the identity of \(G\).

Since \(e_G\) is the identity of \(G\) and \(e_H \in G\), \(e_Ge_H=e_He_G=e_H\).

Further, \(e_H \in H, e_H^{-1}=e_H \in H\).

Since \(e_H, e_H^{-1} \in G\), \(e_He_H^{-1}=e_G\), but \(e_He_H^{-1}=e_H\).  

Hence \(e_G=e_H\).◻

Let \(g,h \in H\).

 

 

Example \(\PageIndex{1}\)

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.

Counter example:

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

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

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

 

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

No.

Counter example:

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.

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

 

Example \(\PageIndex{2}\)

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

Suppose you take this matrix s.t. the trace of the matrix is zero.

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

 

Show that \(H \le G\).

Solution:

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 will show that \( g \ast h^{-1} \in H, \forall \; g,h^{-1} \in H\).

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 \(\PageIndex{3}\)

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 that is a \(2 \times 2\) matrix 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 \(SL_2(\mathbb{R})\). Show that \(SL_2( \mathbb{R}) \le GL_2\mathbb{R})\). 

 

Definition: finite group

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

 

Theorem \(\PageIndex{3}\)

Subgroup test for finite groups:

Let \(H \subseteq G\).

Then \(H \le G\) iff:

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

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

 

Example \(\PageIndex{4}\)

Find all subgroups of  Klein 4 \( (K_4) \) Group \(K_4=\{e, a, b, c\}\) 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 \(K_4.\)

Lattice Diagram of all subgroups

Example \(\PageIndex{5}\)

	\(K_4\)
/	|	\
\(H_1=\{e,a\}\)	\(H_2=\{e,b\}\)	\(H_3=\{e,c\}\)
\	|	/
	\(\{e\}\)

 

 

Special subgroups 

Centralizer of  an element

Definition: Centralizer of an element

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  \(C(a)\le G\).

 

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  \(C(a)\le G\).

Example \(\PageIndex{6}\)

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

Theorem \(\PageIndex{4}\)

 

Let \((G, \ast )\) 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 \((x_1x_2^{-1})g=x_1gx_2^{-1}\).

     \(=(x_1g)x_2^{-1}\)

     \(=(gx_1)x_2^{-1}\)

     \(=g(x_1x_2^{-1})\).◻

 

Note

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

 

Example \(\PageIndex{7}\)

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

 Solution:

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

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

Example \(\PageIndex{8}\)

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 \(C(a) \subseteq C(a^{-1})\), \(C(a) = C(a^{-1})\).◻

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 \( g_1h=hg_1\), \(g_2h=hg_2\), \(\forall g \in H\).

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 of \(G\).

Conjugacy classes of \(G\) are subsets of \(G\) in which elements are conjugate to each other. 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 \(G\).◻