2.3: Subgroups
- Page ID
- 131257
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Subgroups are non-empty subsets of a group that are themselves groups. In this section, we will explore how to define and characterize subgroups.
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).
To ensure a set is a group, we need to check for the following five conditions:
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Non-empty set,
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Closure,
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Associativity,
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Identity, and
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inverse.
The following theorem allows us to check three conditions (rather than 5) to ensure a subset is a subgroup.
- The trivial subgroups of a group \(G\) are \(\{e\}\} and \(G.\)
- \((\mathbb{Z},+)\) is a subgroup of \((\mathbb{R},+).\) In fact \(\mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}\) are subgroups of \(\mathbb{R}\) under addition.
- But \((\mathbb{Q}^*,\cdot)\) is not a subgroup of \((\mathbb{R},+).\) (why?)
Let \((G, \ast )\) be a group then \(H \subseteq G\) is a subgroup if and only if
The following conditions are satisfied:
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The identity \(e \in G\) is in \(H\).
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If \(h_1,h_2 \in H\), then \(h_1 \ast h_2 \in H\).
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If \(h \in H\) then \(h^{-1} \in H\).
- Proof
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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 \(H \le G\).◻
-
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From (2), we have shown \(h_1 \ast h_2 \in H, \forall h_1,h_2 \in H\).
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From (1), we have shown the identity.
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From (3), we have shown the inverses exist.
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Every element of \(h_1,h_2,h_3 \in G\) therefore associativity applies because \(H\) inherits it from \(G\) by definition.
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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:
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,\(h_1 \ast h_2 \in H, \forall h_1,h_2 \in H\)
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\(e \in H\), where \(e \in G\),
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\(h^{-1} \in H, \forall \; h \in H\),
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\(h_1(h_2h_3)=(h_1h_2)h_3, \forall \; a,b,c \in H\).
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The following theorem allows us to check two conditions (rather than 5) to ensure a subset is a subgroup.
Let \((G, \ast)\) be a group.
Then \(H (\subseteq G)\) is a subgroup of \(G\) iff the following conditions are satisfied:
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\(e_G \in H\), and
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\(g \ast h^{-1} \in H, \forall \; g,h \in H\).
- Proof:
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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 \(h_1,h_2,h_3 \in H\).
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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 \(g,h \in H\).
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Consider \((\mathbb{R}^*, \bullet)\). Decide if the following sets form subgroups of \(\mathbb{R}^*\).
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\(H=\{x\in \mathbb{R}^*|x=1\) 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, \(H \not \le \mathbb{R}^*\).◻
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\(K = \{x \in \mathbb{R}^*| x \ge 1\}\).
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 \(a^{-1} \not \exists \forall a \in K\), \(K \not \le \mathbb{R}^*\).◻
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 \(H \le G\).
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 \(SL_2( \mathbb{R}) \le GL_2\mathbb{R})\).
Let \(n\) be a nonnegative integer, Then show that \(n\mathbb{Z}=\{nm|m\in \mathbb{Z}\} is a subgroup of \((\mathbb{Z},+).\)
- Answer
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Add texts here. Do not delete this text first.
Let \(G\) be a group. If \(G\) is finite and \(|G|=n\) then we say that the order of \(G=n\).
Subgroup test for finite groups:
Let \(H \subseteq G\).
Then \(H \le G\) iff:
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\(H \ne \{\}\). (non-empty).
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\(a, b \in H\) then \(ab \in H\) (closed).
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
\(K_4\) | ||
/ | | | \ |
\(H_1=\{e,a\}\) | \(H_2=\{e,b\}\) | \(H_3=\{e,c\}\) |
\ | | | / |
\(\{e\}\) |
Special subgroups
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 \(C(a)\le G\).
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 \(C(a)\le G\).
Let \(G=GL(2,\mathbb{R})\). Then find
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\(C \Bigg( \begin{bmatrix}1 & 1\\
1 & 0 \end{bmatrix}\Bigg) \). -
\(C \Bigg( \begin{bmatrix} 0 & 1\\
1 & 0 \end{bmatrix}\Bigg)\).
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 \(C\Bigg(\begin{bmatrix} 1 & 1\\
1 & 0 \end{bmatrix}\Bigg) =\Bigg\{A= \begin{bmatrix} a & b\\
b & a-b \end{bmatrix} \Bigg| a,b \in \mathbb{R}, \det(A) \ne 0 \; \Bigg\}\).
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\(C \Bigg( \begin{bmatrix} 0 & 1\\
1 & 0 \end{bmatrix}\Bigg)\).
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
Let \(G\) be a group. Then the center of \(G\) is defined as, \(Z(G)=\{x \in G| gx=xg \forall \; g \in G \}.\)
Let \((G, \ast )\) be a group then \(Z(G)\) is a subgroup of \(G.\)
- Proof:
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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 that if \(G\) is abelian, \(Z(G)=G\).
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, \(\mathbb{Z}(G) =\Bigg\{ \begin{bmatrix} a & 0\\
0 & a \end{bmatrix} \Bigg| a \in \mathbb{R}, a \ne 0 \; \Bigg\}\).
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
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 \(C(H)=G\).
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 \(G\).
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 \(G\), \(N(H)= \{ xHx^{-1} | x \in G\}.\)
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 \(G\).◻