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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\} $$

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Remember that the determinant is linear in each of the rows. If you start with a matrix in $GL_2$, can you modify it slightly to get something with determinant $1$? –  Dylan Moreland Aug 25 '11 at 7:43
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3 Answers 3

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.

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$|SL(2,\mathbb{F_p})\cap H_2|=(p-1)p$, 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.

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The answer suggested by Alex B. is more useful(general): one can ask same question with $F_p$ replaced by any field (infinite field also); then the answer of Alex B. is valid; whereas my answer and that of Robinson necessarily uses finiteness of groups. –  David Aug 25 '11 at 9:06
True, but the Frattini argument adapts to other contexts, including this one. In the algebraic group situation, the unipotent radical of the Borel can play the role of the Sylow, and the Frattini argument works with respect to any normal subgroup containing it. –  Geoff Robinson Aug 25 '11 at 9:27
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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.

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