MHB Order of product of elements in a group

[Arnold1]

Hello.

I'm just beginning my course in algebra. I've been reading Milne, Group Theory ( http://www.jmilne.org/math/CourseNotes/GT310.pdf page 29).
I've found there a very nice proof of the fact that given two elements in a finite group, we cannot really say very much about their product's order. However, there are some things about the proof I do not quite understand. Namely - the first paragraph. What are the images of elements in $$SL_2(\mathbb{F}_q)/ \{+-I\}$$ and why do we divide the orders of $$a, \ b, \ c$$ by $$2$$?

Is it because the centre($$ \{+-I\} $$) has order $$2$$ and thus by Lagrange's theorem, the order of the quotient group must be two times smaller?I would really appreciate a thorough explanation. Maybe you know a simpler proof of the fact (about the order of product of elements)?Thank you.

 

[Opalg]

Hello.

I'm just beginning my course in algebra. I've been reading Milne, Group Theory ( http://www.jmilne.org/math/CourseNotes/GT310.pdf page 29).
I've found there a very nice proof of the fact that given two elements in a finite group, we cannot really say very much about their product's order. However, there are some things about the proof I do not quite understand. Namely - the first paragraph. What are the images of elements in $$SL_2(\mathbb{F}_q)/ \{+-I\}$$ and why do we divide the orders of $$a, \ b, \ c$$ by $$2$$?

If $a$ has order $2m$ then $a^m$ has order $2$. But, as Milne points out, $-I$ is the unique element of order $2$ in $\text{SL}_2(\mathbb{F}_q)$. Therefore $a^m = -I$, so that (the coset of) $a^m$ is the identity element in the quotient group $\text{SL}_2(\mathbb{F}_q)/\{\pm I\}.$ It follows that the image of $a$ has order $m$ in the quotient group.

Thank you for that link! It looks as though Milne's notes are an excellent free online resource for group theory.

 

[Arnold1]

You're welcome. Thank you for the explanation.

 

[I like Serena]

Welcome to MHB, Arnold! :)

What are the images of elements in $$SL_2(\mathbb{F}_q)/ \{+-I\}$$

Let's first start with the elements.
The elements have the form {a,-a}, where $a \in SL_2(\mathbb{F}_q)$, which is a 2x2 matrix with elements from $\mathbb Z/q\mathbb Z$.

There is a so called natural or canonical function $SL_2(\mathbb{F}_q) \to SL_2(\mathbb{F}_q)/ \{\pm I\}$, given by $a \mapsto \{a,-a\}$.
Milne means that the image of an element a is {a,-a}, since a itself is not an element of the quotient group.

why do we divide the orders of $$a, \ b, \ c$$ by $$2$$?

Let's pick an example in $SL_2(F_3)$
$$a=\begin{pmatrix}1 & 1 \\ 1 & 2\end{pmatrix},\ a^2 = \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$$
So $a$ has order 4.
Since $a^2 = -I$ already belongs to the coset $\{\pm I\}$, which is the identity element, the order of {a,-a} is 2.

Is it because the centre($$ \{+-I\} $$) has order $$2$$ and thus by Lagrange's theorem, the order of the quotient group must be two times smaller?

Yes.

 

[johng1]

Hi,
Here's some unsolicited advice. From your question, I don't think you should be studying Milne's monograph yet. Case in point, the theorem in your question. The first assertion is that SL(2,q) with q=p^k has a unique element of order 2. I think this requires some proof, which to me is not obvious. Furthermore, the statement is false if p=2 -- If you can easily prove the above, I retract my advice. So my advice to you is to get a good grounding in basic algebra before you go back to Milne.

P.S. In case you can read German, the very old but still very good book Endliche Gruppe I by B. Huppert is a very comprehensive treatment. It is virtually self contained; that is, all necessary facts from other branches of algebra are presented.