Order of product of elements in a group

Arnold

New member
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 $$\displaystyle SL_2(\mathbb{F}_q)/ \{+-I\}$$ and why do we divide the orders of $$\displaystyle a, \ b, \ c$$ by $$\displaystyle 2$$?

Is it because the centre($$\displaystyle \{+-I\}$$) has order $$\displaystyle 2$$ 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

MHB Oldtimer
Staff member
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 $$\displaystyle SL_2(\mathbb{F}_q)/ \{+-I\}$$ and why do we divide the orders of $$\displaystyle a, \ b, \ c$$ by $$\displaystyle 2$$?
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.

Last edited:

Arnold

New member
You're welcome. Thank you for the explanation.

Klaas van Aarsen

MHB Seeker
Staff member
Welcome to MHB, Arnold!

What are the images of elements in $$\displaystyle SL_2(\mathbb{F}_q)/ \{+-I\}$$
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 $$\displaystyle a, \ b, \ c$$ by $$\displaystyle 2$$?
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($$\displaystyle \{+-I\}$$) has order $$\displaystyle 2$$ and thus by Lagrange's theorem, the order of the quotient group must be two times smaller?
Yes.

johng

Well-known member
MHB Math Helper
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=pk 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.