Proof verification: Prove whether or not every proper subgroup of a nonabelian group is nonabelian.

Question

Prove whether or not every proper subgroup of a nonabelian group is nonabelian.

Proof: Counterexample: Let $G=GL_{2}(\mathbb{R})$ be the general linear group of $2\times 2$ matrices and let $H=\left \{ \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix},\begin{bmatrix} 2 &0 \\ 0& 1 \end{bmatrix} \right \}$. Then $H$ is a proper subgroup of the nonabelian group $G$, yet $H$ is abelian. $\square$

Answer 1

Obviously the trivial subgroup is abelian no matter what $G$ is.

But there's more: in fact every group (abelian or not) has a nontrivial abelian subgroup. Indeed, let $G$ be a group and $g\in G$. Obviously $\langle g\rangle$ is cyclic hence abelian. It is nontrivial if $g$ is. It is proper if $G$ is not of prime order (note that in the case $G$ is of prime order then $\langle g\rangle=G$ hence $G$ is cyclic and abelian).

Answer 2

This is not correct: your $H$ is not a subgroup! For instance, writing $A=\begin{bmatrix} 2 &0 \\ 0& 1 \end{bmatrix}$, $A\in H$ but $A\cdot A=\begin{bmatrix} 4 &0 \\ 0& 1 \end{bmatrix}\not\in H$. (However, there is an even easier example that does work. Is there some subset of your $H$ which is a subgroup?)

Answer 3

Consider the dihedral group $D_n$ of order $m=2n$; the symmetry group of the regular $n$-gon. The subgroup $Z_n $, the group of rotations is a cyclic subgroup and therefore abelian .