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I am really sorry that I am not able to solve this one, thank you for your help.

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    If 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

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$a$ 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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    @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