# How to find the order of SL(2,4)?

• Feb 22nd 2009, 07:07 AM
Amanda1990
How to find the order of SL(2,4)?
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).
• Feb 22nd 2009, 07:30 AM
ThePerfectHacker
Quote:

Originally Posted by Amanda1990
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)?

You should know that $\displaystyle |\text{GL}(n,q)| = (q^n-1)(q^n - q)...(q^n - q^{n-1})$.
You should also know $\displaystyle \text{GL}(n,q)/\text{SL}(n,q) \simeq K^{\times}$ where $\displaystyle |K^{\times}| = q-1$, now you can solve for $\displaystyle |\text{SL}(n,q)|$.
• Feb 22nd 2009, 10:45 AM
Amanda1990
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...
• Feb 22nd 2009, 10:51 AM
ThePerfectHacker
Quote:

Originally Posted by Amanda1990
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.

Do you remember the proof behind Lagrange's theorem? In the proof it is shown that if $\displaystyle r$ is the number of left cosets of $\displaystyle B$ in $\displaystyle A$ then $\displaystyle r|B| = |A| \implies r = |A|/|B|$. However, if $\displaystyle A$ is a normal subgroup then $\displaystyle r = |A/B|$ and so $\displaystyle |A/B| = |A|/|B|$.

Quote:

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 $\displaystyle kI$ where $\displaystyle I$ is the identity matrix? If so then $\displaystyle \det (kI) = k^2$ and we need this determinant to be $\displaystyle 1$, by definition of special linear groups. Thus, $\displaystyle k = \pm 1$.
• Feb 22nd 2009, 12:19 PM
Amanda1990
Great response again. Surely $\displaystyle |A/B| = |A|/|B|$ [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 $\displaystyle k^2 = 1$ 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 $\displaystyle k^2 = 1$, 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.
• Feb 22nd 2009, 01:03 PM
ThePerfectHacker
Quote:

Originally Posted by Amanda1990
Great response again. Surely $\displaystyle |A/B| = |A|/|B|$ [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).

If $\displaystyle B$ is not a normal subgroup of $\displaystyle A$ then you cannot form $\displaystyle A/B$!
However, you can still talk about $\displaystyle (A:B)$, the number of left cosets of $\displaystyle B$ in $\displaystyle A$.
Yes, $\displaystyle (A:B) = |A|/|B|$.

Quote:

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 $\displaystyle k^2 = 1$ implies k = 1 or -1, but actually this is of course clear. Thanks.
If $\displaystyle k^2 = 1 \implies k^2 - 1 = 0\implies (k-1)(k+1) = 0 \implies k=1,-1$.
Of course if field charachteristic is two then there is only one solution since $\displaystyle 1=-1$.