How do you find the order of a special linear group? For example, how do you find order of SL(2,4)? I think the answer's 60 but really would like a proof. How do you find the order of SL(2,5)?
Finally, here's a problem set on one of my problem sheets - "Show that the only scalar matrices in SL(2,5) are I and -I. This seems fairly obvious but I'm not sure how to begin a proof. And why do you think this question was asked? (No proof required, I just want to know the reason behind asking this).
OK thanks, this is really helpful. I have not been given the results you quoted but was able to derive them. I've got two quick questions:
1) If A is a group and B is a subgroup then is the order of the quotient group A / B equal to the order of A divided by the order of B? ie is |A/B| = |A| / |B|? I appreciate the answer is probably blindingly obvious, but I've thought about it a bit and just ended up confusing myself, so an answer and quick justification would be good here.
2) Why should the only scalar matrices in SL(2,5) be I or -I? I originally thought this looked obvious but actually now I'm not so sure...
I am not sure what you mean by "scalar matrices", do you mean where is the identity matrix? If so then and we need this determinant to be , by definition of special linear groups. Thus, .2) Why should the only scalar matrices in SL(2,5) be I or -I? I originally thought this looked obvious but actually now I'm not so sure...
Great response again. Surely [COLOR=Black]holds irrespective of whether B is a normal subgroup of A or just any subgroup of A? (I assume you mean B is a normal subgroup and not A is a normal subgroup).
This is indeed what I meant by a scalar matrix - I was just slightly worried that in a general field it may not be obvious that implies k = 1 or -1, but actually this is of course clear. Thanks.
EDIT: Actually I'm back to thinking that this last point may not be so obvious - what is clear is that 1 and -1 satisfy , but how do we know there are no other such elements in a general field satisfying this? (Above I was thinking about the Fundamental Theorem of Algebra and concluded that there were only two roots but this kind of argument won't work in general fields)
EDIT 2: This can be done by using the fact that all finite fields of the same order are isomorphic.
However, you can still talk about , the number of left cosets of in .
If .This is indeed what I meant by a scalar matrix - I was just slightly worried that in a general field it may not be obvious that implies k = 1 or -1, but actually this is of course clear. Thanks.
Of course if field charachteristic is two then there is only one solution since .