I am really sorry that I am not able to solve this one, thank you for your help.
a question on normal subgroup of $GL_n(\mathbb{C})$ and $GL_n(\mathbb{R})$
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0Do you have any suspicions? Which statements are problematic? – 2012-07-19
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0regarding $b$ and $c$ I am totally stuck, I am trying now for $a$ by hand – 2012-07-19
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0If you just look at $a$ for a quick second, if you let $AB = \{ ab \, | \, a \in A, b \in B\}$ with $A, B \subset G$, then $H\{g\} = \{g\} H$ by normality and $H\{g\} H = (H \{g\}) H$, i.e. the operation I just defined is associative. Then $a$ just says that $HgH = (Hg)H = (gH)H = g(HH) = gH$, so in this frame of mind it's not that hard. By hand, i.e. by writing $HgK = \{ \dots \}$, it's not that hard too. – 2012-07-19
1 Answers
$a$ and $c$ are true by abstract group theory.
The first one shoudn't be too hard. Just write it out.
The third one basically says that the path component of identity in a topological group is a normal subgroup. To show it, for arbitrary $\varphi:[0,1]\to G$, $a\in G$, consider $\varphi_a(x)=axa^{-1}$.
The second one isn't true. To see this notice that conjugation in general linear group is just a change of basis, so \begin{pmatrix}1 &1\\ 0 &1\end{pmatrix} and \begin{pmatrix}1 &0\\ 1 &1\end{pmatrix} are clearly conjugate.
Edit: I just noticed that the hint I've dropped for c only shows how to see that $H$ is normal, not that it is actually a subgroup. For that, take $a,b\in H$ corresponding to paths $\varphi,\psi$ and consider $\varphi\cdot \psi$ (pointwise multiplication) and for each $a\in H$ consider $\varphi^{-1}$ (not the inverse function, but the pointwise inverse to $\varphi$).
This proof shows us something slightly stronger: if $G$ is an arbitrary group, and $N\lhd G$ arbitrary, then the union of path components of all elements of $N$ is again a normal subgroup. In this case we have $N$ as the trivial subgroup, which is of course always normal.
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0Found the same example and was about to write it, but I'll leave the answer to you because I didn't even take time to think about c. Good answer, +1 – 2012-07-19
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0i do not understand $(b)$, let H be the set of all invertible upper triangular matrcies with complex entries, to show H is not normal subgroup of $GL_n(\mathbb{C})$,we need to show $ghg^{-1}$ does not belongs to H for some $g\in GL_n(\mathbb{C})$, could you tell me what is your $g$ and $h$? – 2012-07-20
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0I also dont understand $c$, to prove H is a subgroup,let $a,b\in H$ then take the path $\phi_a$ joining $a$ and $I$ and $\psi$ joining $b$ and $I$, are you saying $\phi\times \psi$ will be a path joining $a$ and $b$? – 2012-07-20
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0@Patience: for b) i've given you $h$ and $ghg^{-1}$ and told you how to see that they're conjugate. It looks like a homework, so you should figure some of it yourself. For c: not $\varphi\times \psi$, but $\varphi\cdot\psi$. If $\varphi(0)=a$ and $\psi(0)=b$ and $\varphi(1)=\psi(1)=e$, then $\varphi\cdot\psi(0)=\ldots$, while $\varphi\cdot\psi(1)=\ldots$? – 2012-07-20
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0@tomasz: Could you please take a look at [this question](http://math.stackexchange.com/q/284159/11619). I hazard a guess that the OP is Polish, and she seems to have problems with English terminology. May you can assist? – 2013-01-22