Finding the size of the smallest subgroup of $\mathrm{GL}(2,\mathbb{R})$ containing two matrices.

I've only been exposed to basic abstract algebra (Like Definition of a group + Subgroup lemma etc) and some first year linear algebra. (We have not seen lagranges theorem, incase that is required for this question).

I was hoping if someone could show an elementary way of doing this question:

Let $$H$$ be the smallest subgroup of $$GL(2,\mathbb{R})$$ containing both $$A = \pmatrix{0 & 1 \\ -1 & 0} \text{ and } B =\pmatrix{0 & 1 \\ 1 & 0}.$$ Show that $$H$$ has eight elements. (Recall $$GL(2,\mathbb{R})$$ is the group of $$2\times 2$$ invertible matrices with real entries under matrix multiplication)

Is there a way of doing the question without making a long 8 by 8 multiplication table? (That was my initial attempt, but it was far too tedious).

Thanks!

• Calculate $A^2, A^3, A^4...$, you will end up at the identity at some point, which says something about the inverses of those matrices. Then form the products with the other matrix. It does not take much time. It is perhaps helpful to think about what the second matrix does multiplied from the right or left to another matrix. – idle mathematician Jun 9 '19 at 23:47
• @B.Swan I got $A^4 = I$, implying $(A^2)^{-1} = A^2$? Sorry I'm not sure how this might help? – user523384 Jun 9 '19 at 23:52
• Are you sure you typed the matrix right? I get different results – idle mathematician Jun 10 '19 at 0:01
• Oh, my apologies! Really sorry there's an extra "1" there – user523384 Jun 10 '19 at 0:02
• I was wondering why I was having order of $A$ as $6$. – ArsenBerk Jun 10 '19 at 0:03

First, verify that $$|A|=4$$ and $$|B|=2$$.
Next compute $$AB,A^2B,A^3B,BA,$$ and so on.
But from here you will get that $$BA=A^3B$$ By using this relation, we obtain that $$BA^2=A^2B, BA^3=AB$$ By using these relation, every element in $$H$$ can be written as $$A^iB^j$$ where $$0\le i\le3$$ and $$0\le j \le1$$.
So we get $$H=\{1,A,A^2,A^3,B,AB,A^2B,A^3B\}$$ and this is actually isomorphic to $$D_8$$.

• Ah this makes sense, Just to confirm: - For $A^{i}B^{j}$ is $0\le i \le 3$ and $0 \le j \le 1$ because $A^4 = I$ and $B^2=I$ so any higher power can be reduced to those? - How did you get $|A| = 4$ and $|B| = 2$? It doesn't seem like the determinant nor the number of elements? Thanks – user523384 Jun 10 '19 at 1:58
• @user523384 For your first question, that is true because of three things: $A^4=1,B^2=1,BA=A^3B$ because the third relation allows you to "swap" the position of B with A. Just for an exercise you may try with $BABA^3B$. For second question, $|A|$ denote the order of $A$, which is the smallest positive integer $n$ such that $A^n=i$. In group theory to avoid confusion, the determinant normally will be denoted by $det(A)$. – Alan Wang Jun 10 '19 at 2:02
• This is helpful! Thank you. For the exercise, $BABA^3B = A^6B = A^2 B \in H$, right? Once we have established $A^iB^j$ where $0\le i\le3$ and $0\le j \le1$, we just test the possible combinations of $A$ and $B$ which are unique? Also for the existence of an inverse, would $B^{2-j}A^{4-i}$ be the inverse? How would you show this is in the set $H$? Sorry for so many questions! And thank you for your excellent answers :) – user523384 Jun 10 '19 at 2:11
• @user523384 Yes your answer is correct. Secondly you can list all the eight elements and check that they are all distinct. For their inverse, you still can use the three main relations $$A^4=1,B^2=1,BA=A^3B$$ to tackle it. Because $A^4=1, B^2=1$, we have $A^{-1}=A^3$ and $B^{-1}=B$. So for example, $(A^3B)^{-1}=B^{-1}A^{-3}=BA^9=BA=A^3B$ which is still in $H$. – Alan Wang Jun 10 '19 at 2:17
• @user523384 I do not think that express an inverse for general powers is a good way because it will involve modulo operation. Also check the inverse of only $AB,A^2B,A^3B$ will be enough. – Alan Wang Jun 10 '19 at 2:27

Every element of $$\langle A,B\rangle$$ can be written as a word $$A^{\alpha_1}B^{\beta_1}A^{\alpha_2}B^{\beta_2}\dots A^{\alpha_k}B^{\beta_k}$$ for some $$k$$, with $$0\le\alpha_j\le3$$ and $$0\le\beta_i\le1$$. Using the commutation relation $$BA=A^3B$$, we can put each element in the form $$A^{\alpha}B^{\beta}$$, where again $$0\le\alpha\le3$$ and $$0\le\beta\le1$$. So there are at most $$8$$ elements.

Next check that $$I,A,A^2,A^3,B,A^3B,A^2B$$ and $$AB$$ are all distinct.

• Thank you! Just to confirm, is $0 \le \alpha \le 3$ and $0 \le \beta \le 1$ because any higher powers can be reduced using $A^4 = I = B^2$? – user523384 Jun 10 '19 at 2:00
• Correct. That's the reason. – Chris Custer Jun 10 '19 at 2:27

I don't know you will find this one short or not but it is elementary I can say.

Say $$A = \pmatrix{0 & 1 \\ -1 & 0}$$ and $$B= \pmatrix{0 & 1 \\ 1 & 0}$$. Then, first we can multiply and element with itself until we get identity matrix $$I$$. For $$B$$, we have $$B^2 = I$$. For $$A$$, we have $$A^2 = \pmatrix{-1 & 0 \\ 0 & -1}$$ so $$A^2 \in H$$. Then, $$A^3 = \pmatrix{0 & -1 \\ 1 & 0}$$ so $$A^3 \in H$$. Then, $$A^4 = I$$. Now, we need to check $$AB$$, $$A^2B, A^3B$$. Can you take it from here?

Note that this procedure comes from the closure of groups.

• Thank you for this, this makes sense, but I'm actually sure when to "stop" or to realise that we "now have a group". Would you have me do this in a table until every product is closed? – user523384 Jun 10 '19 at 0:07
• There are a few things to note here: 1. $A$ is cyclic, meaning that $A^k=I$ for some natural $k$. So the inverses of powers of $A$ are powers of $A$, so you dont need to calcuate them by hand. 2. $B^2=I$, so the inverse of say $AB$ will be $BA^{k-1}$, also no need to find them by hand. Next you just do $A, A^2, A^3, A^4$, then $BA, AB, A^2B, BA^2, ABA, BAB, A^3B, BA^3$ and here we notice something.. – idle mathematician Jun 10 '19 at 0:15
• Yes, B.Swan concluded it actually. – ArsenBerk Jun 10 '19 at 0:23
• Does ( AB = -BA ) assist in any way? – user523384 Jun 10 '19 at 0:29
• You can write $-$ as $-I$, which is actually $A^2$. So we actually have $AB = A^2BA$. You are on the right track. – ArsenBerk Jun 10 '19 at 0:30