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2.2: Properties of Group Elements

  • Page ID
    131256
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    Theorem \(\PageIndex{1}\)

    Let \((G, \ast)\) be a group.    Then

    1. The identity is unique,

    2. For each \(a \in G\), one and only one inverse exists.

    3. For each \(a \in G, \ (a^{-1})^{-1}=a\).

    4. \((a\ast b)^{-1}=b^{-1} \ast a^{-1}, \forall a,b \in G.\)

    Note if \(G\) is abelian,\((a\ast b)^{-1}=b^{-1} \ast a^{-1}=a^{-1} \ast b^{-1}, \forall a,b \in G\).

    5. If \(xy=xz, \forall x,y,z \in G\), then  \(y=z\).  (Left cancellation)

    6. If \(yx=zx, \forall x,y,z \in G\), then  \(y=z\).  (Right cancellation)

     

    Answer

    Proof:

    Let \((G,\ast)\) be a group.

    1. We shall show that identity is unique.

    Assume that \(G\) has two identity elements, \(e_1\) and \(e_2\).

    Thus \(a \ast e_1=e_1 \ast a=a\) and \(a \ast e_2=e_2 \ast a=a\), \(\forall a \in G\).

    We will show that \(e_1=e_2\).

    Consider \(e_1 \ast e_2=e_2 \ast e_1 =e_1\) and \(e_2 \ast e_1=e_1 \ast e_2 =e_2\).

    Thus since \(e_1=e_2\), the identity is unique.◻

     

    2. We shall show that for each \(a \in G\), one and only one inverse exists.

    Let \(a \in G\).

    Assume that \(a\) has two inverses, \(b\) and \(c\).

    Then \(a \ast b=b \ast a=e\) and \(a \ast c=c \ast a=e\).

    We shall show that \(b=c\).

    Consider \(b=b \ast e\)

          \(=b \ast(a \ast c)\)

          \(=(b \ast a) \ast c\)

          \(=e \ast c\)

          \(=c\).

    Thus \(b=c\) and for each \(a \in G\), there exists one and only one inverse.

    3. We shall show that for each \(a \in G, \ (a^{-1})^{-1}=a\).

    Let \(a \in G\). Then  \(a \ast a^{-1}=a^{-1}\ast a=e\).

    Thus \((a^{-1})^{-1}=b^{-1}=a\).

    Assume that \(G\) has two identity elements, \(e_1\) and \(e_2\).

    Thus \(a \ast e_1=e_1 \ast a=a\) and \(a \ast e_2=e_2 \ast a=a\), \(\forall a \in G\).

    We will show that \(e_1=e_2\).

    Consider \(e_1 \ast e_2=e_2 \ast e_1 =e_1\) and \(e_2 \ast e_1=e_1 \ast e_2 =e_2\).

    Thus since \(e_1=e_2\), the identity is unique.◻

    4. 

    Let \(a,b \in G\).

    Then \(a \ast b \in G\), \(a^{-1} \in G\), \(b^{-1} \in G\) and \(a \ast b \in G\).

    Consider \((b^{-1} \ast a^{-1})(a \ast b)= b^{-1} \ast (a^{-1}a) \ast b)\)

            \(=b^{-1}\ast e \ast b\)

            \(=b^{-1}\ast b\)

            \(=e\).

    And consider  \((a \ast b)(b^{-1} \ast a^{-1})=a \ast ( b^{-1} \ast b) \ast a^{-1}\)

       \(=a^{-1}\ast e \ast a\)

               \(=a^{-1}\ast a\)

       \(=e\).

    Since \((b^{-1} \ast a^{-1})(a \ast b)=e= (a \ast b)(b^{-1} \ast a^{-1})\), then \((a\ast b)^{-1}=b^{-1} \ast a^{-1}, \forall a,b \in G\).◻

    5.

    Let \(x,y,z \in G\) s.t. \(xy=xz\).

    Note:  \(x^{-1} \in G\) since \(G\) is a group.

    Thus \(x^{-1}(xy)=x^{-1}(xz)\)

    Thus \((x^{-1}x)y=(x^{-1}x)z\) since  associative.

    Thus \(ey=ez\) and \(y=z\).◻

    6. 

    Let \(x,y,z \in G\) s.t. \(y \ast x=z \ast x\).

    Note:  \(x^{-1} \in G\) since \(G\) is a group.

    Thus \((y \ast x) \ast x^{-1}=(z \ast x) \ast x^{-1}\)

    Thus \(y \ast (x^{-1} \ast x)=z \ast (x^{-1} \ast x)\) since  associative.

    Thus \(y \ast e=z \ast e\) and \(y=z\).◻

     

    Theorem \(\PageIndex{2}\)

    Let \(G\) be a group.

    Let \(g,h \in G\) and \(m,n \in \mathbb{N}\).

    Then

    1. \(g^m \ast g^n=g^{m+n}\) where \(g^m=g \ast \cdots \ast g\).  Note there would be \(m\) \(g\)’s.

    2. \((g^m)^n=g^{mn}\).

    3. \((gh)^{-m}=(h^{-1} g^{-1})^m\).   

    Note

    With the addition operation \(g^m=mg.\) 

    Example \(\PageIndex{1}\)

    Let \(G\) be a group and suppose that \((ab)^2=a^2b^2\) for all \(a\) and \(b\) in \(G\).  Prove that \(G\) is an abelian group.

    Solution

    Screen Shot 2023-06-29 at 1.54.57 PM.png

    Let \(G\) be a group.

    Let \((ab)^2=a^2b^2, \; \forall a,b \in G\).

    We shall show that \(G\) is abelian.

    Let \(a,b \in G\).

    We shall show that \(ab=ba\).

    Note that \((a \star b)\star (a \star b)= (a \star a) \star (b \star b)\).

    Consider that \(a^{-1} \star (a \star b) \star (a \star b)\star b^{-1}=a^{-1} \star (a \star a) \star (b \star b) \star b^{-1}\).

    Then \((a^{-1} \star a) \star b \star a \star (b \star b^{-1})=(a^{-1} \star a)\star a \star b \star (b \star b^{-1})\).

    Thus \(e \star b \star a \star e=e \star a \star b \star e\).

    Thus \(b \star a=a \star b\).◻

    Note:  Operator * included for students.

     


    This page titled 2.2: Properties of Group Elements 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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