Prove commutative subgroup

Question

Let $K$ be the group consisting of all invertible $2\times2$ matrices with coefficients in $\mathbb Z/3\mathbb Z$. Here the group is the usual matrix multiplication, and the unit element is $e:=\begin{bmatrix}1&0\\0&1\end{bmatrix} $

Consider the matrices $p:=\begin{bmatrix}0&1\\1&1\end{bmatrix} $

$l:=\begin{bmatrix}1&1\\0&2\end{bmatrix} $

Let $R:=\{h^ng^m : m,n \in \mathbb{Z}_{>0} \}$ and $F:=\{g\in G:\det(g)= 1 \pmod{3}\}$ are subgroups of $G$.

Show that $R \cap F$ is a commutative group.

I have shown that it is a subgroup, but I don't know how to show that for all $x,y\in(J\cap H)$ we have $xy=yx$.

Could somebody please help me? Thank you!

Answer 1

Hint: it is enough to find a generating set for $J\cap H$ and to show that its elements commute. Three generators should suffice.

Answer 2

Let $F=\Bbb F_3$ be the field with three elements. Then $$ \begin{aligned} G &= \operatorname{GL}(2,F) \ ,\qquad &|G| &= (3^2-1)(3^2-3)=48\ ,\\ H &= \operatorname{SL}(2,F) \ ,\qquad &|H| &= |G/\pm1|=|G|/2 =24\ ,\\ <g> &= \{1,g,g^2,\dots, g^7\}\ & g&\text{ has order }8\ ,\\ <h> &= \{1,h\}\ & h&\text{ has order }2\ ,\\ hgh^{-1} &= g^3\ ,\\ J:=<g,h> &=\{1,g,g^2,\dots, g^7,\ h, hg, hg^2,\dots, hg^7\}\ & |J|&=16\ ,\\ \det g=\det h&=-1\ ,\\ J\cap H &=\{1,g^2,g^4, g^6,\ hg, hg^3,hg^5, hg^7\}\ , &|J\cap H|&=8\ ,\\ &=<g^2,hg>\ , \end{aligned} $$ and we check the commutativity of the above two generators of $J\cap H$, using computations with final results in the above explicit list of elements, we use the relations $hgh^{-1} = g^3$ and $h^2=1$, so $hgh=g^3$, and $gh=hg^3$: $$ \begin{aligned} hg\cdot g^2 &= hg^3\ ,\\ g^2\cdot hg &=g\ gh\ g=g\ hg^3\ g=gh\ g^4=hg^3\ g^4\\ &= hg^7\ . \end{aligned} $$ The group $J\cap H$ with eight elements is thus not commutative.

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Computer check, here sage:

sage: F = GF(3)    # the field with three elements
sage: G = GL(2, F)
sage: H = SL(2, F)
sage: g = matrix( F, 2,2, [0,1,1,1] )
sage: g.multiplicative_order()
8
sage: h = matrix( F, 2,2, [1,1,0,2] )
sage: h.multiplicative_order()
2
sage: h*g*h == g^3
True
sage: J = G.subgroup( [g, h] )
sage: J.order()
16
sage: J_cap_H = G.subgroup( [g^2, h*g] )
sage: J_cap_H.order()
8
sage: [ j.matrix().det() for j in J_cap_H ]
[1, 1, 1, 1, 1, 1, 1, 1]
sage: J_cap_H.is_commutative()
False
sage: h*g * g^2 == g^2 * h*g
False