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I'm trying to solve the following problem:

Let $G$ be the group of matrices of the form $$\begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix}$$ where $a \in (\mathbb{Z}/p)^*$ and $b \in \mathbb{Z}/p$. Describe all normal subgroups of $G$.

My partial solution is as follows,

A normal subgroup is the union of conjugacy classes. Each conjugacy class consists of elements of the form $$\begin{bmatrix} x & y \\ 0 & 1 \end{bmatrix}\begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix}\begin{bmatrix} x & y \\ 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} x & y \\ 0 & 1 \end{bmatrix}\begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix}\begin{bmatrix} \frac1x & \frac{-y}{x} \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & (1 - a)y + bx \\ 0 & 1 \end{bmatrix}$$

where $a$ ranges across all elements in $(\mathbb{Z}/p)^*$ and $b$ ranges across all elements in $\mathbb{Z}/p$.

From here we have to take unions and I feel like there should be a more elegant way of describing this but I'm sure.

Math_Day
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    Hint: The conjugacy classes of $G$ are represented by $\begin{pmatrix}a & b \ 0 & 1\end{pmatrix}\ \sim\ \begin{pmatrix}a & 0 \ 0 & 1\end{pmatrix}$ when $a\neq 1$, $\begin{pmatrix}1 & b \ 0 & 1\end{pmatrix}\ \sim\ \begin{pmatrix}1 & 1 \ 0 & 1\end{pmatrix}$ when $b\neq 0$, and of course $\begin{pmatrix}1 & 0 \ 0 & 1\end{pmatrix}$. Using this, classify the normal subgroups of $G$ according to their image under the homomorphic surjection of $G$ unto $(\mathbb{Z}/p)^*$ via the determinant. It helps to note that the nonempty fibers must all be of the same cardinality (i.e., that of the kernel). – Rafi Apr 18 '23 at 04:14

1 Answers1

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HINT:

  1. First note the equation $$\det(x)\det(y)\det(x^{-1}) = \det(y).$$ So in particular, the set of $p$ matrices $y$ satisfying $\det(y)=1$ form a normal subgroup in $G$. So equivalently, the set $G(1) = \Big\{ \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix}\Big\}$ form a normal group in $G$.

  2. In fact, in light of 1.. let $Z^*$ be any proper multiplicative subgroup of $(\mathbb{Z}/p\mathbb{Z})^*$. Then the set $G(Z^*)=\Big\{ \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix}; a \in Z^*\Big\}$ $=\{x \in G; \det(x) \in Z^*\}$ also form a normal group in $G$.

  3. Finally, the identity element $e$ forms by itself a normal subgroup of $G$, and of course $G$ is itself a normal subgroup of itself.

So now use the work you have done in your OP to show that those are the only normal subgroups of $G$. Indeed, let $H$ be any other normal subgroup of $G$. Suppose that $H$ is not as prescribed by 1.,2., or 3. as above. Then as $G(1)$ as in 1. above has $p$ elements [with $p$ prime], it follows that either $H$ contains $G(1)$ as a subgroup or $H \cap G(1) =\{e\}$. In any event, lest $H$ not be as in 1. or 2., there must then be a matrix $x \in H$ such that $\det(x) \not = 1$. However, as you have shown yourself OP: if $x$ and $x'$ are $2$ elements in $G$ with the same determinant not equal to $1$, then $x$ and $x'$ are conjugate to each other. So then $H$ contains all matrices $x' \in G$ satisfying $\det(x')=\det(x)$. Furthermore, as $H$ must also contain $x^r; r$ a positive integer; it follows that $H$ must also contain all matrices $y$ satisfying $\det(y)=\det^r(x)$ for some positive integer $r$. But then this implies that $H$ must be as in 2. or $H$ must be the entire group $G$, and thus indeed, 1.,2, and 3. characterize all the possible normal subgroups of $G$.

Mike
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