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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 $\det(a)$.

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