Catalin wrote: Let $n \geq 4$ and $G$ be a subgroup of $GL_2(\mathbb{C})$ with $|G| = n.$ Prove that there exists $H \le G$ such that $\{ I_2 \} \neq H$ and $H \neq G$ such that $XYX^{-1} \in H, \: \forall X \in G$ and $\forall Y \in H$ I don't see how that could be true. What if $G$ has no non-trivial subgroup? For example, given a prime number $p > 3,$ let $\alpha:= e^{\frac{2\pi i}{p}}$ and then let $G$ be the subgroup of $\text{GL}(2,\mathbb{C})$ generated by $\begin{pmatrix}\alpha & 0 \\ 0 & 1\end{pmatrix}.$ Then $G$ is a group of order $p$ and so it has no non-trivial subgroup.

There's a mistake in the problem; n must be an even number.

ysharifi wrote: Catalin wrote: Let $n \geq 4$ and $G$ be a subgroup of $GL_2(\mathbb{C})$ with $|G| = n.$ Prove that there exists $H \le G$ such that $\{ I_2 \} \neq H$ and $H \neq G$ such that $XYX^{-1} \in H, \: \forall X \in G$ and $\forall Y \in H$ I don't see how that could be true. What if $G$ has no non-trivial subgroup? For example, given a prime number $p > 3,$ let $\alpha:= e^{\frac{2\pi i}{p}}$ and then let $G$ be the subgroup of $\text{GL}(2,\mathbb{C})$ generated by $\begin{pmatrix}\alpha & 0 \\ 0 & 1\end{pmatrix}.$ Then $G$ is a group of order $p$ and so it has no non-trivial subgroup. I'm sorry, I forgot to add the fact that $n$ is even.

Catalin wrote: I'm sorry, I forgot to add the fact that $n$ is even. OK, that makes sense now. So the problem is to show that $G$ has a non-trivial normal subgroup. First notice that we are done if $G$ is abelian because then every subgroup of $G$ is normal and so, for example, we can choose $H=\langle Y \rangle,$ where $Y$ is an element of order two (clearly $H \ne G$ because $|G| > 2.$) So suppose, to the contrary, that $G$ is non-abelian and $G$ has no non-trivial normal subgroup. Consider the group homomorphism $f: G \to \mathbb{C}^{\times}$ defined by $f(X)=\det(X).$ Since $\ker f$ is a normal subgroup of $G,$ we must have $\ker f=\{I_2\}$ or $\ker f = G.$ We cannot have $\ker f =\{I_2\}$ because then $G$ would be embedded into $\mathbb{C}^{\times}$ making $G$ abelian. So $\ker f=G,$ i.e. $$\det X=1$$for all $X \in G.$ Now let $Y \in G$ be an element of order two. So $Y^2=I_2,$ which implies that $Y$ is diagonalizable. So $$Y=P\begin{pmatrix}a & 0 \\ 0 & b \end{pmatrix}P^{-1}$$for some $P \in \text{GL}(2,\mathbb{C}).$ Since $Y^2=I_2,$ the set of eigenvalues of $Y,$ which is $\{a,b\},$ is a subset of $\{-1,1\}.$ Since $\det Y=1,$ we can either have $a=b=1$ or $a=b=-1.$ So either $Y=I_2$ or $Y=-I_2.$ Since $Y$ has order two, $Y \ne I_2$ and hence $Y=-I_2.$ Clearly $H:=\langle Y\rangle$ is a non-trivial normal subgroup of order two in $G$ because $Y=-I_2$ is a central element of $G.$ This is a contradiction!