## walters Group Title 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 one year ago one year ago

1. terenzreignz Group Title

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

2. walters Group Title

A=|dw:1362839192398:dw|

3. terenzreignz Group Title

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 Group Title

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 Group Title

looks like we could take out i's

6. walters Group Title

|dw:1362839499204:dw|

7. terenzreignz Group Title

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 Group Title

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

9. terenzreignz Group Title

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 Group Title

are we not suppose to use those axioms of subgroups

11. terenzreignz Group Title

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 Group Title

Shall we proceed? :)

13. walters Group Title

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

14. terenzreignz Group Title

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 Group Title

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

16. experimentX Group Title

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

17. terenzreignz Group Title

Yes. I'm very happy :D

18. experimentX Group Title

I love identity matrix :DDD

19. terenzreignz Group Title

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

20. walters Group Title

yes

21. walters Group Title

we can proceed

22. terenzreignz Group Title

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 Group Title

yes

24. terenzreignz Group Title

So now, we compute A^4

25. terenzreignz Group Title

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

26. terenzreignz Group Title

See where it has taken us?

27. terenzreignz Group Title

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

28. walters Group Title

yes it is the identity

29. terenzreignz Group Title

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 Group Title

wow this is interesting

31. terenzreignz Group Title

So... have you arrived at your conclusion?

32. walters Group Title

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

33. terenzreignz Group Title

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 Group Title

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

35. terenzreignz Group Title

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

36. terenzreignz Group Title

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