Set of invertible elements in Banach algebra












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Let $A$ be a Banach algebra, $G(A)$ set of invertible elements from A and $G_0(A)$ connected component of $G(A)$ that contains 1. Prove that $G_0(A)$ is a normal subgroup of $G(A)$.



I have proved that $G(A)$ is a group, $G(A)$ is open subset of $A$ and $B(e;1) subseteq G_0(A)$.
$B(e;1) subseteq G_0(A)$ follows from $$(e-x)^{-1} = sum_{n=0}^{infty} x^n $$
But I don't know how to prove that $G_0(A)$ is a group, let alone normal subgroup.










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  • 3




    $begingroup$
    Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
    $endgroup$
    – KCd
    Dec 6 '18 at 23:13
















1












$begingroup$


Let $A$ be a Banach algebra, $G(A)$ set of invertible elements from A and $G_0(A)$ connected component of $G(A)$ that contains 1. Prove that $G_0(A)$ is a normal subgroup of $G(A)$.



I have proved that $G(A)$ is a group, $G(A)$ is open subset of $A$ and $B(e;1) subseteq G_0(A)$.
$B(e;1) subseteq G_0(A)$ follows from $$(e-x)^{-1} = sum_{n=0}^{infty} x^n $$
But I don't know how to prove that $G_0(A)$ is a group, let alone normal subgroup.










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
    $endgroup$
    – KCd
    Dec 6 '18 at 23:13














1












1








1


1



$begingroup$


Let $A$ be a Banach algebra, $G(A)$ set of invertible elements from A and $G_0(A)$ connected component of $G(A)$ that contains 1. Prove that $G_0(A)$ is a normal subgroup of $G(A)$.



I have proved that $G(A)$ is a group, $G(A)$ is open subset of $A$ and $B(e;1) subseteq G_0(A)$.
$B(e;1) subseteq G_0(A)$ follows from $$(e-x)^{-1} = sum_{n=0}^{infty} x^n $$
But I don't know how to prove that $G_0(A)$ is a group, let alone normal subgroup.










share|cite|improve this question











$endgroup$




Let $A$ be a Banach algebra, $G(A)$ set of invertible elements from A and $G_0(A)$ connected component of $G(A)$ that contains 1. Prove that $G_0(A)$ is a normal subgroup of $G(A)$.



I have proved that $G(A)$ is a group, $G(A)$ is open subset of $A$ and $B(e;1) subseteq G_0(A)$.
$B(e;1) subseteq G_0(A)$ follows from $$(e-x)^{-1} = sum_{n=0}^{infty} x^n $$
But I don't know how to prove that $G_0(A)$ is a group, let alone normal subgroup.







abstract-algebra functional-analysis






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edited Dec 6 '18 at 23:26







ig97

















asked Dec 6 '18 at 23:08









ig97ig97

155




155








  • 3




    $begingroup$
    Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
    $endgroup$
    – KCd
    Dec 6 '18 at 23:13














  • 3




    $begingroup$
    Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
    $endgroup$
    – KCd
    Dec 6 '18 at 23:13








3




3




$begingroup$
Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
$endgroup$
– KCd
Dec 6 '18 at 23:13




$begingroup$
Can you tell us what you have thought of so far, rather than just state a problem ("Prove that...") and nothing more?
$endgroup$
– KCd
Dec 6 '18 at 23:13










1 Answer
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$begingroup$

This is true for any topological group. $h_g(x) =gxg^{-1}$ is continuous, so $h_g( G_0)$ is connected and contained in $G_0$ since it contains $1$, since $h_g(1)=1$, so it is $G_0$.






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    1












    $begingroup$

    This is true for any topological group. $h_g(x) =gxg^{-1}$ is continuous, so $h_g( G_0)$ is connected and contained in $G_0$ since it contains $1$, since $h_g(1)=1$, so it is $G_0$.






    share|cite|improve this answer









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      1












      $begingroup$

      This is true for any topological group. $h_g(x) =gxg^{-1}$ is continuous, so $h_g( G_0)$ is connected and contained in $G_0$ since it contains $1$, since $h_g(1)=1$, so it is $G_0$.






      share|cite|improve this answer









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        1












        1








        1





        $begingroup$

        This is true for any topological group. $h_g(x) =gxg^{-1}$ is continuous, so $h_g( G_0)$ is connected and contained in $G_0$ since it contains $1$, since $h_g(1)=1$, so it is $G_0$.






        share|cite|improve this answer









        $endgroup$



        This is true for any topological group. $h_g(x) =gxg^{-1}$ is continuous, so $h_g( G_0)$ is connected and contained in $G_0$ since it contains $1$, since $h_g(1)=1$, so it is $G_0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 23:22









        Tsemo AristideTsemo Aristide

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        56.8k11444






























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