Let $$A=\begin{pmatrix}i & 0\\ 0 & -i\end{pmatrix}$$ and

$$B=\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}.$$

Let $Q=\langle A,B\rangle.$

Prove that Q has an automorphism of order 3.

  • $\begingroup$ Not much really, the definition I have of automorphism I find vague. An isomorphism with G = G∗ is said to be an automorphism of G . $\endgroup$ – ZZS14 Oct 31 '13 at 9:28
  • $\begingroup$ what definition do you have? $\endgroup$ – user87543 Oct 31 '13 at 9:29
  • $\begingroup$ basic thing you should do is to find orders of $A$ and $B$ $\endgroup$ – user87543 Oct 31 '13 at 9:30
  • $\begingroup$ I have done that. Sorry, the orders are both 4. $\endgroup$ – ZZS14 Oct 31 '13 at 9:31
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    $\begingroup$ I don't really understand what the link is saying, sorry $\endgroup$ – ZZS14 Oct 31 '13 at 10:09

As Praphulla Koushik has noted in another answer, the key is to recognize this as the quaternion group: ignoring the usual convention that $I$ denotes the identity matrix and instead calling the identity '$\mathbb{1}$', and changing our variable names from $A$ and $B$ to the suggestive $I$ and $J$, we have $I^2 = \begin{pmatrix}-1 & 0\\ 0 & -1\end{pmatrix} = -\mathbb{1}$, where of course $(-\mathbb{1})^2=\mathbb{1}$; likewise $J^2 = -\mathbb{1}$. And if we define the matrix $K$ by $K=IJ=\begin{pmatrix}0 & i\\ i & 0\end{pmatrix}$, then we also have $K^2=-\mathbb{1}$; in other words, the three matrices $I, J, K$ satisfy the relations $I^2=J^2=K^2=IJK=-\mathbb{1}$. (Note that the last one derives trivially by expanding one instance of $K=IJ$ in $K^2=-\mathbb{1}$). Furthermore, we have $JKI=-\mathbb{1}$ and $KIJ=-\mathbb{1}$ (you can prove the last two by algebraic manipulations using the relations you already have, without doing any matrix arithmetic; this is a good exercise). A complete list of all the elements in the matrix group would be $\mathbb{1}, -\mathbb{1}, I, J, K, -I, -J, -K$ (where e.g. $-I$ is $-\mathbb{1}\cdot I$) — prove this!

Now, you should see a certain symmetry in the relations among $I, J, K$ that were written above; you should be able to exploit this symmetry to find a morphism $\eta$ by picking suitable 'target' matrices for $\eta(I)$, $\eta(J)$ and $\eta(K)$ and then showing that all of the relations are invariant under application of the morphism $\eta$.

  • $\begingroup$ I have as far as Q={1, A, -A, B, -B, AB, -AB} which is equivalent to what you have, but I am having trouble seeing the symmetry and progressing further than here. Thank you. $\endgroup$ – ZZS14 Oct 31 '13 at 18:56
  • $\begingroup$ @Emily Try rewriting 'AB' as 'C' - having three distinct generators (with a relation between them) is much easier here than thinking in terms of two. A big hint: what differentiates 'A' from 'B' (or in my formulation, 'I' from 'J') in the generating relations? $\endgroup$ – Steven Stadnicki Oct 31 '13 at 19:11
  • $\begingroup$ when you say differentiates A from B, what do you mean? All I can see is that A^2=B^2=-1? $\endgroup$ – ZZS14 Oct 31 '13 at 21:42
  • $\begingroup$ @Emily That's exactly it (and don't forget the 'ABC=-1' relation); there's no difference whatsoever between A and B, inherently. You can't just swap A and B (i.e., via a morphism $\eta$ with $\eta(A)=B, \eta(B)=A, \eta(C)=C$) because (a) that would be order 2, rather than order (3), and (b) the relation $ABC=-1$ isn't invariant; you get $\eta(ABC) = \eta(A)\eta(B)\eta(C) = BAC = 1\neq \eta(-1)$. But consider setting $\eta(A)=B, \eta(B)=C, \eta(C)=A$... $\endgroup$ – Steven Stadnicki Oct 31 '13 at 22:14
  • $\begingroup$ If I have n(A)=B, n(B)=C, n(C)=A, n(ABC)=n(A)n(B)n(C)=BCA= 1 or have I done something wrong $\endgroup$ – ZZS14 Oct 31 '13 at 22:41

first of all i am sorry for making you to do so much non sense.

First of all i believe that you could recognize your $Q$ to be $Q_8$.

So, you have $Q\cong Q_8=\{\pm 1, \pm i, \pm j, \pm ij\}$.

Now, you need to see if there is a possibility for an automoprphism of $Q_8$ to have order $3$

By an "automorphism of a group" you mean a "homomorphism which is bijective"

So, your required map $\eta : Q\rightarrow Q$ has to be a homomorphism.

For that you need $\eta(a.b)=\eta(a).\eta(b) \forall a,b\in Q$

$\eta(a)=\eta(a.1)=\eta(a).\eta(1)$.Thus, $\eta(1)=1$.

Now, you have $1=\eta(1)=\eta(-1.-1)=\eta(-1).\eta(-1)$ i.e., $(\eta(-1))^2=1$

So, only possible choice of $\eta(-1)$ for non trivial bijective function $\eta$ is $-1$

i.e., $\eta(-1)=-1$

You have $Q=\{\pm1,\pm i,\pm j,\pm ij\}$

Now, let us go for next generator that is $i\in Q_8$

you have six possibilities for $i$ namely $\{\pm i, \pm j, \pm k\}$

you want your map to be of order $3$ so you should not send $i$ to $i$.

with previous messed up post you could be able to see that i can not send $i$ to $-i$

Let us consider a map

$\eta : Q\rightarrow Q $ which send $i\rightarrow j$

please see that this should automatically fix image of $i$ i.e.,


Now let us try(cry) for possible image of $j$

For similar reason as in previous messed up (and also for the reason that we need it to be bijective),

I can not send $j$ to $j$ or even $-j$

So, possible choices for image of $j$ are $\{\pm i, \pm ij\}$

Suppose $j$ goes to $i$ Then, we would have

$$i \rightarrow j\\-i\rightarrow -j\\ j\rightarrow i (\text{just now defined})\\-j\rightarrow -i(\text{as $j$ is fixed, so is $-j$})$$

It is upto you to see where does $ij$ and $-ij$ goes to,

just use that $\eta$ is a homomorphism i.e., if $a$ and $b$ are fixed in domain of $\eta$, then $\eta(ab)$ is fixed as $\eta(ab)=\eta(a)\eta(b)$

So, you should be able to see that

$$i \rightarrow j\\-i\rightarrow -j\\ j\rightarrow i \\-j\rightarrow -i\\ij\rightarrow ji=-ij \\ ji\rightarrow -ji=ij$$

For similar reason why $i$ should not be sent to $-i$ you should be able to see $ij$ should not be sent to $-ji$. thus above map is not what we need.

So, next possible choice is $j\rightarrow -i$. I would request you to check yourself that this would then be of order $4$

So, next possibility for image of $j$ is $ij$.

Now, you should be able to see where does other elements should be mapped to namely,

$$ i \rightarrow j\\-i\rightarrow -j\\ j\rightarrow ij \\-j\rightarrow -ij \\ij\rightarrow jij=i \\-ij \rightarrow -i$$

Now, you do not have to check this is a homomorphism because it is defined keeping in mind of homomorphism condition.

You can see clearly that this is a bijection.

So, this is an automorphism.

Coming to the order, we see that :

$$i \rightarrow j\rightarrow ij\rightarrow i\\ -i \rightarrow -j\rightarrow -ij\rightarrow -i\\ j\rightarrow ij\rightarrow i \rightarrow j\\ij\rightarrow i \rightarrow j\rightarrow ij$$

Please check other images to make sure it is of order $3$.

I hope this should work.

  • $\begingroup$ I'm confused when it comes to applying the definition of homomorphism $\endgroup$ – ZZS14 Oct 31 '13 at 14:23
  • $\begingroup$ I can see why n(1)=1, but not really why the same holds for -1 $\endgroup$ – ZZS14 Oct 31 '13 at 14:57
  • $\begingroup$ would I be right in saying -n(-a)=n(a)? if so is n(-1)=-1 because n(a)=n(-a.-1)=n(-a).n(-a) $\endgroup$ – ZZS14 Oct 31 '13 at 15:09
  • $\begingroup$ your justification for $\eta(-a)$ is not correct.. please see the edit $\endgroup$ – user87543 Oct 31 '13 at 17:21
  • $\begingroup$ Yes, have just seen it thank you. i'm not really sure how to show that i to -i is a homomorphism. $\endgroup$ – ZZS14 Oct 31 '13 at 17:30

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