# Subgroup of order $4$ of matrix group

Let $G$ be the set of all real $\ 2\times 2$ matrices $\begin{pmatrix} a & b \\ 0 & d \end{pmatrix}$ where $ad\neq 0$. We know that $G$ forms a group under matrix multiplication. Construct in the $G$ a subgroup of order $4$.

My solution: Let $H\subset G$ be the following set $H=\{E,-E,A,-A\}$ where $E=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $A=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$. It's easy to check that $A^2=E$ and $A^{-1}=A$. The set $G$ holds the property of binary operation, existence of inverse and identity elements and the property of associativity.

Hence the set $H$ is the subgroup of $G$ and $|H|=4$.

Right?

Let me ask one question: For the binary operation should we check out all possibillities by hand? It's annoying :)

• Are $a,b,d\in \mathbb{F}_q$ for a prime power $q$? For $q=2$ we only get the trivial group. Nov 23, 2017 at 12:27
• @DietrichBurde, your comments is always difficult :) I do not know what is $\mathbb{F}_q$
– RFZ
Nov 23, 2017 at 12:30
• $\mathbb{F_q}$ is a field with $q$ elements, example: en.wikipedia.org/wiki/GF(2), is $\mathbb{F_2}$ Nov 23, 2017 at 12:39
• @DietrichBurde,So what? We are talking $\mathbb{R}$
– RFZ
Nov 23, 2017 at 12:40
• Yes, now it says real:) Nov 23, 2017 at 13:22

Your reasoning is correct, since the group $G$ consists of real matrices.

• ..... and that is a reason for downvote ?? A usual style in Austria ?
– Fred
Nov 23, 2017 at 13:43
• Dietrich, dann entschuldige ich mich !
– Fred
Nov 23, 2017 at 14:11

Your reasoning is correct except in the case where the field from which you take $a, b, d$ has characteristic $2$; in that case you have $1 = -1$, and all your matrices $E, -E, A, -A$ are identical.

• I do not know what is the characteristic of the field. I am in the begining of group theory from the book of Herstein.
– RFZ
Nov 23, 2017 at 12:35
• @RFZ: Okay, but even in that book the author should say a few things about what $a, b, d$ are, right? Nov 23, 2017 at 12:38
• $a,b,d$ are real numbers
– RFZ
Nov 23, 2017 at 12:39
• and the characteristic of real numbers is $0$
– RFZ
Nov 23, 2017 at 12:39
• @RFZ: Oops, you're right; it says so in the OP; I had missed that - sorry. In that case your argument is right (since $1\ne -1$). Nov 23, 2017 at 12:41