Question

Question based on (Groups)

Answer 1

jjj let the mapping \[\tau _{ab}\] for a,b element R ,maps the reals by the ruel
\[\tau _{ab}:x \rightarrow ax+b.Let G={\tau _{ab}:a \neq 0}\]
Determine whether or not G is an abelian group under the composition of mappings

Answer 2

oops ignore jjj

Answer 3

Well, let \(\tau_{ab},\tau_{cd}\in G\). Then \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b.\]Also, \[\tau_{cd}\circ\tau_{ab}(x)=\tau_{cd}(ax+b)=cax+bc+d.\]Since \(acx+ad+b\neq acx+bc+d\), it must be that \(G\) is not abelian.

Answer 4

am i not suppose to check whether is a group using the condition before i can conclude that is an abelian group

Answer 5

It just asked if it was an abelian group. Since the elements don't commute, it can't be an abelian group even if it is a group. However, if we want to check it's a group, we can do that as well.

Have you shown any of the axioms for proving it's a group yet?

Answer 6

i am failing to show the axioms

Answer 7

First, we can see that \(G\) is closed under composition. \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b=\tau_{(ac)(ad+b)}.\]We also have the identity \(\tau_{10}\). Now we just have to check associativity and inverses.

Answer 8

why do u chose to use

Answer 9

"\(\circ\)" is a relatively common notation to denote the composition of two functions. You can also use \(\tau_{ab}(\tau_{cd}(x))\) is you prefer.

Answer 10

can u also use * if u wnt

Answer 11

For inverses, we want \[\tau_{ab}\circ\tau_{cd}(x)=acx+ad+b=x\]for some \(c,d\in\mathbb{R}\). We can see that \(c=a^{-1}\) and \(d=a^{-1}(-b)\) work for this direction. For the other, \[\tau_{cd}\circ\tau_{ab}(x)=cax+bc+d=a^{-1}ax+ba^{-1}-a^{-1}b=x,\]so we do indeed have inverses.

Answer 12

If you make it clear your group function is composition of functions, you can use * as well.

Answer 13

As for associativity, I'm just going to use the fact that composition of functions is always associative. If you have not heard of that yet, we can still go over how to prove this particular function is associative.

Answer 14

please do

Answer 15

let \(\tau_{ab},\tau_{cd},\tau_{ef}\in G\). Then\[\tau_{ab}\circ(\tau_{cd}\circ\tau_{ef})(x)=\tau_{ab}=\tau_{ab}(cex+cf+d)=acex+acf+ad+b\]and\[(\tau_{ab}\circ\tau_{cd})\circ\tau_{ef}(x)=(\tau_{(ac)(ad+b)})(ex+f)=acex+acf+ad+b\]Since these are equal, it's associative, and we have a group.

Answer 16

k it is a group but not an abelian group

Answer 17

Yup.

Answer 18

thnx

Answer 19

You're welcome.

Answer 20

can we verify identity

Answer 21

\[\tau_{10}(x)=1x+0=x\]