Group generated by two subgroups

Question

What is the subgroup of $GL_{2}(\mathbb{F}_{p})$ that is generated by $ $$SL_{2}(\mathbb{F}_{p})$ and $H_{2}$ where $$ H_{2}=\left\{ \left(\begin{array}{cc} a & b\\ 0 & c\end{array}\right):a,c\in\mathbb{F}_{p}^{\times},b\in\mathbb{F}_{p}\right\} $$

Answer 1

Your $H_2$ is traditionally denoted by $B$ and is called a Borel subgroup (sometimes the standard Borel subgroup) of $GL_2(\mathbb{F}_p)$

Exercise: $GL_2(\mathbb{F}_p)$ is a disjoint union $B\sqcup BwB$, where $w=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, and $BwB=\{b_1wb_2|b_1,b_2\in B\}$.

This fact is fairly important in the representation theory of $GL_2(\mathbb{F}_p)$, and among other things it should immediately answer your question.

Answer 2

$|GL(2,\mathbb{F_p})|=(p^2-1)(p^2-p)$,

$|SL(2,\mathbb{F_p})|=(p^2-1)(p^2-p)/(p-1)$,

$|H_2|=(p-1)^2p$.

$|SL(2,\mathbb{F_p})\cap H_2|=(p-1)p$, since it consists of elements of $H_2$ with $c=a^{-1}$

Then $| \langle SL(2,\mathbb{F_p}),H_2\rangle| \geq|SL(2,\mathbb{F_p}) H_2|$

$= |SL(2,\mathbb{F_p})|\cdot|H_2|/|SL(2,\mathbb{F_p})\cap H_2| = (p^2-1)(p^2-p)=|GL(2,\mathbb{F_p})|$

So the subroup is in fact whole group.

Answer 3

This is a good opportunity to make use of the Frattini argument. The Frattini argument use Sylow's Theorem to factorize a finite group $G$ in the presence of a normal subgroup $N.$ It states that if $N \lhd G$ and $P \in {\rm Syl}_{p}(N),$ then we have $G = NN_{G}(P).$ The proof is not difficult: for any $g \in G,$ we have $P^{g} \leq N$ as $N \lhd G.$ But $P^g$ is another Sylow $p$-subgroup of $N,$ so $P^{g} = P^{n}$ for some $n \in N.$ Then $gn^{-1} \in N_{G}(P),$ so that $g \in N_{G}(P)N = NN_{G}(P).$ Since $g$ was arbitrary, we have $G = NN_{G}(P).$

In your case, $P$, the group of upper unitriangulr matrices (ie upper triangular with $1$'s on the diagonal) is in ${\rm SL}(2,\mathbb{F}_p)$ and is a Sylow $p$-subgroup of that group. Its normalizer (in ${\rm GL}$) is precisely the group you call $H_2.$ Thus we have ${\rm GL}(2,\mathbb{F}_p) = {\rm SL}(2,\mathbb{F}_p)H_2$ by a Frattini argument.