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.

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?)

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 $\PageIndex{2}$

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.

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 $\PageIndex{3}$

 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 $\PageIndex{4}$

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 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 $SL_2( \mathbb{R}) \le GL_2\mathbb{R})$. 

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

Add texts here. Do not delete this text first.

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 $\PageIndex{6}$

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 $\PageIndex{7}$: Lattice diagram of subgroups of $K_4$

 

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 $G$,  $C(a)=\{g \in G|ag=ga\}$. 

 

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

Example $\PageIndex{8}$

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 $\PageIndex{9}$

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 $\PageIndex{10}$

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 $g_1h=hg_1$, $g_2h=hg_2$, $\forall h \in H$.

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 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 $G$.◻