## walters one year ago Let G be the group of all invertible 2 x 2 matrics over the C of complex numbers with matrix multiplication a)Find the subgroup of G generated by

1. terenzreignz

This matrix. I got you the first time :) $\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]$

2. walters

A=|dw:1362839192398:dw|

3. terenzreignz

I have a better handwriting!!! Just kidding. So, we're taking a cyclic group, generated by that matrix above, using matrix multiplication. Evaluate this...$\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]$

4. terenzreignz

If matrices make you dizzy, (they make me dizzy) use this formula... $\huge \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}p & q \\ r & s\end{matrix}\right]=\left[\begin{matrix}ap+br & aq+bs \\ cp+dr & cq+ds\end{matrix}\right]$

5. experimentX

looks like we could take out i's

6. walters

|dw:1362839499204:dw|

7. terenzreignz

And of course, I was doing it inefficiently :D Bloody matrices... Thanks @experimentX @walters As you 'heard', we have this convenient setup... $\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=i\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]$

8. walters

before going too far wat if i use A and other matrix that in invertible

9. terenzreignz

Now THAT's going too far :) We're only asked to figure out what the elements of the cyclic group generate by $\huge\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]$

10. walters

are we not suppose to use those axioms of subgroups

11. terenzreignz

You can do that, but why bother? Let G be a group. For any element g in G, <g> (the set generated by repeatedly operating g on itself) is always a subgroup of G. In other words, the cyclic group generated by an element of a group is a subgroup :)

12. terenzreignz

Shall we proceed? :)

13. walters

yes, it seems i have no idea "how to do it"

14. terenzreignz

Okay, you can verify this, but $\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]$

15. terenzreignz

You know what? For convenience, we can let $\huge A=\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]$

16. experimentX

isn't it obvious when you get negative of identity matrix?? when you get that matrix ... you should be very happy.

17. terenzreignz

Yes. I'm very happy :D

18. experimentX

I love identity matrix :DDD

19. terenzreignz

So, @walters You're good so far with $\huge A^2=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]$?

20. walters

yes

21. walters

we can proceed

22. terenzreignz

So, now, let's work on A^3... We get... $\huge A^3 = A^2A=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=\left[\begin{matrix}0 & -i \\ -i & 0\end{matrix}\right]$ Following this so far?

23. walters

yes

24. terenzreignz

So now, we compute A^4

25. terenzreignz

$\large A^4=A^2A^2=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]$

26. terenzreignz

See where it has taken us?

27. terenzreignz

So, A^4 it seems, is the identity matrix.

28. walters

yes it is the identity

29. terenzreignz

So... we started with A, and it turns out A^4 is the identity. So that must mean <A> has four elements, right?

30. walters

wow this is interesting

31. terenzreignz

So... have you arrived at your conclusion?

32. walters

no it is just that i have flash even if i don't know wat to do next

33. terenzreignz

Well, the 4 elements are just $\huge <A>=\left\{I_2 \ , \ A \ , \ A^2 \ , \ A^3 \right\}$ Where $\huge A=\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]$

34. walters

so it means there is no need to show those axioms of subgroup

35. terenzreignz

No need at all :) It has already been assumed (correctly) that <A> is a subgroup.

36. terenzreignz

Well, I'm sleepy, but I think we covered the gist of this question :) Signing off now :D ---------------------------------- Terence out