Please Login to access more options.

Problem 75 (The Set Product $Ha$ Preserves Determinants)

Let $m=2$ and let $p$ be a prime (it could be any integer, but $U(p)$ is easier to understand when $p$ is prime). Let $H=\text{SL}(2,\mathbb{Z}_p)$ and $G=\text{GL}(2,\mathbb{Z}_p)$. Let $a\in G$, so $a$ is a 2 by 2 invertible matrix. Recall that the set $Ha$ is the set $$Ha=\{ha\mid h\in H\},$$ so to obtain an element in $Ha$, we just take a matrix $h$ with determinant 1 and then multiply it on the right by the matrix $a$.

  1. Let $b\in GL(2,\mathbb{Z}_p)$. Prove that $b\in Ha$ if and only if $\det(b)=\det(a)$.
    • When you prove if $b\in Ha$ then $\det(b)=\det(a)$, you'll show that every matrix in $Ha$ has the same determinant as $a$.
    • When you prove if $\det(b)=\det(a)$ then $b\in Ha$, you'll show that every matrix with the same determinant as $a$ must be in $Ha.$
  2. Prove that the function $f:H\to Ha$ defined by $f(h)=ha$ is a bijection. (What is the inverse of multiplying on the right by $a$?)
  3. Explain why $|H|=|Ha|$, in other words the number of matrices whose determinant equals 1 is the same as the number of matrices whose determinant equals $\det(a)$.

Solution

Let $p$ be prime. Let $H=SL(2,\mathbb{Z}_p)$ and $G=GL(2,\mathbb{Z}_p)$.

1. Let $b\in G$. We will prove $b\in Ha$ if and only if $\det(b)=\det(a)$. Suppose $b\in Ha$. We will show $\det(b)=\det(a)$. Since $b\in Ha$, we know $b=ha$ for some $h\in H$. We now compute

$$ \begin{align} \det(b)&=\det(ha)\\ &=\det(h)\det(a)\\ &=(1)\det(a)\\ &=\det(a) \end{align} $$

Now suppose $\det(b)=\det(a)$. We will show $b\in Ha$. Note that $b=be_G=b(a^{-1}a)=(ba^{-1})a$. We will now show that $ba^{-1}\in H$. Since $a^{-1}$ is invertible, and $b$ is invertible, it follows that $ba^{-1}$ is invertible. Note that $\det(a^{-1})=(\det(a))^{-1}$. We now compute

$$ \begin{align} \det(ba^{-1})&=\det(b)\det(a^{-1})\\ &=\det(a)\det(a^{-1})\\ &=\det(a)(\det(a))^{-1}\\ &=1 \end{align} $$

Thus $ba^{-1}\in H$, from which it follows that $b\in Ha$.

2. We will prove that $f$ is a bijection by showing that $f$ has an inverse. I claim that $g: Ha \rightarrow H$ defined by $g(k)=ka^{-1}$ is an inverse of $f$. Let $x\in H$. We compute $g(f(x))=f(x)a^{-1}=(xa)a^{-1}=x(a^{-1}a)=xe_H=x$. Let $y\in Ha$. We compute $f(g(y))=g(y)a=(ya^{-1})a=y(a^{-1}a)=ye_H=y$. Hence $g$ is an inverse of $f$, and we conclude that $f$ is a bijection.

3. Since there exists a bijection from $H$ into $Ha$, we know that $|H|=|Ha|$.

From Josh: Ben, this is not my work. I began a solution but never completed it so I asked Shaughn to write up his solution. I told him to erase mine and write up his own, which he did. This one is his write up. I'm going to submit it again in hopes that this pops up on your end as well. I also changed the tag to his name so he will receive the proper credit and fill the write-up count if he hasn't already.
Tags

Change these as needed.

  • When you are ready to submit this written work for grading, add the phrase [[!Submit]] to your page. This will tell me that you have completed the page (it's past rough draft form, and you believe it is in final draft form). Don't type [[!Submit]] on a rough draft.
  • If I put [[!NeedsWork]] on your page, then your job is to review what I've written, address any comments made, and then delete all the comments I made. When you have finished reviewing your work, leave [[!NeedsWork]] on your page and type [[!Submit]]. (Both tags will show up). This tells me you have addressed the comments.
  • I'll mark your work with [[!Complete]] after you have made appropriate revisions.