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2.3: Subgroups

  • Page ID
    131257
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    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

    Screen Shot 2023-06-29 at 1.48.28 PM.pngLet \(G\) 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:Screen Shot 2023-07-03 at 11.38.16 AM.png

     
    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 \(e_He_H^{-1}=e_H\).  Screen Shot 2023-07-03 at 11.40.19 AM.png

      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, \(H \not \le \mathbb{R}^*\).◻

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

    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 \(\PageIndex{7}\)
      \(K_4\)  
    / | \
    \(H_1=\{e,a\}\) \(H_2=\{e,b\}\) \(H_3=\{e,c\}\)
    \ | /
      \(\{e\}\)  

     

     

    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 \(\PageIndex{8}\)

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

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

    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 \(Z(G)=\{x \in G| gx=xg \; \forall g \in G \}\).Screen Shot 2023-07-03 at 1.24.03 PM.png

     

    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 \((x_1x_2^{-1})g=x_1gx_2^{-1}\).Screen Shot 2023-07-03 at 1.25.40 PM.png

         \(=(x_1g)x_2^{-1}\)

         \(=(gx_1)x_2^{-1}\)

         \(=g(x_1x_2^{-1})\).◻

     

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

    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 \(e \in gHg^{-1}\).Screen Shot 2023-07-03 at 1.40.22 PM.png

     

    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 \(G\).◻

     

     


    This page titled 2.3: Subgroups is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.

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