I was trying to solve this question from a past paper of a national level competitive exam.
(In the following question, "$H\le G$" stands for $H$ is a subgroup of $G$)
Pick out the true statements:
a. Let $H\le G$ and $K\le G$. Now for $g\in G$ define $HgK=\{hgk|h\in H,k\in K\}$.Then if $H$ is normal subgroup of $G$,we have $HgH=gH$
b.The set of all $n\times n$ invertible upper triangular matrices with complex entries is a normal subgroup of $GL_n(\mathbb C)$
c. Identify $M_n(\mathbb R)$ with $\mathbb R^{n^2}$ (with usual topology) and let $G\le GL_n(\mathbb R)$. Define $H=\{A\in G|\exists f:[0,1]\to G$ continuous such that $f(0)=A,f(1)=I_n\}$ Then $H$ is normal in $G$
What I have done so far:
I think a. is true but b. is not.
For a. we see that $h_1gh_2=g(g^{-1}h_1g)h_2\in H \forall h_1,h_2\in gH$since $H$ is normal in $G$.
For b. $\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}^{-1}=$$\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\frac13=$$\frac13\begin{bmatrix} 4 & -2 \\ 2 & -1 \end{bmatrix}$ is not upper triangular
Am I right about the above options? and I am completely stuck in c. I can not find any way to solve.
Please help me. Thnx in advance.
For part a you have only shown one inclusion, namely $HgH \subseteq gH$; for a full answer you should show $gH \subseteq HgH$ (even if it's very easy).
For part b, it's not the product of any matrix $A$ with an upper triangular matrix $T$ that you need to consider, it's the conjugation $ATA^{-1}$. Think about diagonalizable matrices, for example: are they all upper triangular?
For part c, suppose $A \in H$ and $M \in G$. You want to prove $MAM^{-1} \in H$. By definition there's some $f : [0,1] \to G$, $f(0) = A$ and $f(1) = I_n$. Now what about $t \mapsto M \cdot f(t) \cdot M^{-1}$?