1. ## Proof of some groups(please guide me)

Hi I'm new in Group's theory(I'm physicist) so i wanted to know if i was making it right with one proof exercise. (And I don't know how to type equations here lol)

(x,y) in R2 , we define A_1,A_2,A_3,A_4 as:

a)A_1(x,y)=(x,y)
b)A_2(x,y)=(-x,-y)
c)A_3(x,y)=(-x,y)
d)A_4(x,y)=(x,-y)
Show that this 4 operations under the operation of aplying one next to the other form a group. Show explicitly the inverse in every operation.

I'll shown what I understand:
with A_1 next to A_2
1) lock property
Let (x,y) be in R2
So, A_2(A_1(x,y))=(-x,-y) in R2

This is all i have. Please guide me, don't give me the answer. Because i really want understand this. Also I was thinking that maybe I misunderstood the exercise.

2. ## Re: Proof of some groups(please guide me)

The group elements should be the 4 actions you've been given.

You need to show that $A_1$ through $A_4$, taken as a set, satisfy the group properties with respect to multiplication/composition. So composition of all actions with each other must be closed, and associative. You need an (as in exactly 1) identity element. You need to show here both existence and uniqueness of the identity element. Lastly, once you have an identity element you can identify each elements inverse.

If you think about the action each of the $A_1$ through $A_4$ has on a point in $\mathbb{R}^2$ this is not so bad. For instance $A_4$ acts like complex conjugation by reflecting a point about the x-axis. Since $(A_4 )(x,y)=(x,-y)\neq(x,y)$ and $(A_4 \circ A_2)(x,y)=(-x,y)\neq(x,y)$ it must be that $A_4$ is both NOT the identity for this group and $A_4$'s inverse is not $A_2$.

3. ## Re: Proof of some groups(please guide me)

Originally Posted by guitandre95
Hi I'm new in Group's theory(I'm physicist) so i wanted to know if i was making it right with one proof exercise.
(x,y) in R2 , we define A_1,A_2,A_3,A_4 as:
a)A_1(x,y)=(x,y)
b)A_2(x,y)=(-x,-y)
c)A_3(x,y)=(-x,y)
d)A_4(x,y)=(x,-y)
Show that this 4 operations under the operation of aplying one next to the other form a group. Show explicitly the inverse in every operation.
Clearly the operation is that of function composition.
It is clear that $A_1$ is the identity and each element is it own inverse.
$\begin{array}{*{20}{c}} \circ &|&{{A_1}}&{{A_2}}&{{A_3}}&{{A_4}} \\ \hline {{A_1}}&|&{{A_1}}&{{A_2}}&{{A_3}}&{{A_4}} \\ {{A_2}}&|&{{A_2}}&{{A_1}}&{{A_4}}&{{A_3}} \\ {{A_3}}&|&{{A_3}}&{{A_4}}&{{A_1}}&{{A_2}} \\ {{A_4}}&|&{{A_4}}&{{A_3}}&{{A_2}}&{{A_1}} \end{array}$ This is known as the Klein-4 group.