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Let $G$ be a subgroup of $GL_n(\Bbb{R})$. Define $$H = \biggl\{ A \in G \ \biggl| \ \exists \ \varphi:[0,1] \to G \ \text{continuous such that} \ \varphi(0)=A , \ \varphi(1)=I\biggr\}$$ Show that $H$ is a normal subgroup of $G$

In this post, the answers are either too advanced (manifolds) or are considered overkill. I was trying to follow the user jug's hints, but I can't understand how to make use of $\phi_A(x) = AxA^{-1}$. I believe I have the same confusion as the OP -- I want to show that for all $A \in G$ and for all $M \in H$, $AMA^{-1} \in H$ -- but I'm struggling very much connecting the idea of a normal subgroup with a continuous function (or path).

Could someone please expand on the hints provided in the linked post?

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I think the important thing about the map $\phi_A:X \mapsto AXA^{-1}$ is that it is continuous. Then for $B\in H$ and its map $\varphi:[0,1]\rightarrow G$ with $\varphi(0)=B, \varphi(1)=I$, the composite of these two maps $\psi$ is continuous and satisfies $\psi(0)=ABA^{-1}, \psi(1)=AIA^{-1}=I$. Therefore, $ABA^{-1} \in H$. –  PavelC Jun 26 '13 at 6:31
    
If $M \in H$, there is some $\phi$ with the above property. Suppose $A \in G$, let $\eta(t) = A \phi(t) A^{-1}$. Then $\eta(0) = A M A^{-1}$ and $\eta(1) = I$. Hence $A M A^{-1} \in H$. –  copper.hat Jun 26 '13 at 6:32
    
Wait $GL_n(\mathbb{R})$ does have to connected components, and both are path connected, the one is $GL_n^+ (\mathbb{R})$ and the other is $GL_n^- (\mathbb{R})$. (Matrices with positive or with negative Determinant). Via Gauss elemination you can easily proof that $GL_n^+ (\mathbb{R})$ is connected and is maximal so all you need to now is that $GL_n^+$ is a normal subgroup. –  Dominic Michaelis Jun 26 '13 at 6:35
    
@DominicMichaelis, it seems that your «via Gauss elimination you can easily prove [...]» is a bit over the head of the OP, no? –  Mariano Suárez-Alvarez Jun 26 '13 at 6:36
    
@MarianoSuárez-Alvarez Do you think I should expand on it? It would be to long for a comment –  Dominic Michaelis Jun 26 '13 at 6:39

1 Answer 1

up vote 4 down vote accepted

Let $H$ be the path-connected component of $I$ in $G$.

  • First, let us notice that $H$ is not empty.

  • Suppose that $A$ and $B$ are in $H$, so that there are maps $\sigma:I\to G$ and $\tau:I\to G$ such that $\sigma(0)=I$, $\sigma(1)=A$, $\tau(0)=I$ and $\tau(1)=B$. Consider the map $$\lambda:t\in I\longmapsto \sigma(t)\tau(t)^{-1}\in G.$$ You can show that

    • the map $\lambda$ is continuous, and that
    • $\lambda(0)=I$ and $\lambda(1)=AB^{-1}$.

    If follows from this that $AB^{-1}$ is also in $H$.

  • Suppose $A$ is in the path-connected component $H$ of $I$, so that there is a path $\sigma:I\to G$ such that $\sigma(0)=I$ and $\sigma(1)=A$.

    Let now $B\in G$ and consider the map $$\tau:t\in[0,1]\longmapsto B\sigma(t)B^{-1}\in G.$$ You should check that

    • it is continuous,

    • its value at $0$ is $I$, and its value at $1$ is $BAB^{-1}$.

    It follows from all this that $BAB^{-1}$ is also an element of $H$.

Now conclude that $H$ is a normal subgroup.

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Heh. Both :-) Context should be enogh to disambiguate in every case —I'll edit later to fix this. –  Mariano Suárez-Alvarez Jun 26 '13 at 21:36
    
In the first part, to allow for the inverse of $B$, do I need $\tau^{-1}$ to be continuous too or is it enough to say $\tau$ is continuous? Is this equivalent to saying if $B\in H$, we want to show $B^{-1}$ is in $H$? And you don't need to edit the post just to fix $I$, it's clear now. –  AlanH Jun 26 '13 at 22:18
    
The function $A\in G\mapsto A^{-1}\in G$ is continuous,so its composition with $\tau:I\to G$ is also continuous. –  Mariano Suárez-Alvarez Jun 26 '13 at 22:22

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