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(This question is related to the previous post I've posted few hours ago: (Dummit's AA, 1.5, P3) Are these presentations of the Quarternion group equivalent?)

I was trying to prove that the presentation $$\langle i, j \mid i^4 = j^4 = 1, ij = j^3 i\rangle $$

generates Quaternion group. The only thing that I couldn't derive was that $i^2 \neq 1$. I tried to derive a contradiction from the supposition $i^2 =1$ but this didn't give me a contradiction (at least not yet) but rather it brought me to an Abelian group that consists of 4 elements with every element other than 1 having order 2.

I'm not sure where I'm wrong.

Does indeed the above presentation generate an Abelian group with its order 4? (I think this can't happen according to the previous post!!)

Or $i^2 = 1$ is a indeed contradictory?

If the latter is the case, would you please show me the derivation?

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le4m
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  • I don't think you can get a contradiction because other groups (including $i=j=1$, the trivial group) satisfy this description. – hasnohat Jun 20 '13 at 00:36

3 Answers3

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You're confused, I think, about the definition of a presentation of a group. That's fair enough, since the idea is slightly subtle.

Formally, the idea of a presentation is as follows: you define some generators like $i,j$ and then consider all expressions in the free group on those symbols, which are strings like $iij^{-1}iji^{-1}jjij$ and so on. Then you impose any rules which the relations, like $i^4=1$, tell you. So e.g. $iiiijjij\equiv jjij$. You also assume $ii^{-1}=i^{-1}i=1$ and so on.

You're not allowed to assume that any other rules for manipulating these things apply. Then you ask how the resulting set of strings works under concatenation. This defines unambiguously the group multiplication.

So for example $\left<i|i^n=1\right>$ is a presentation of the cyclic group of order $n$. The trivial group, even though it satisfies without contradiction the relations, is not defined by this presentation, because you cannot use the relations given to deduce $i^1=1$.

Edit: As you say, the idea is that $i^2=1$ in the group defined by the above cyclic relation ($n>2$) holds $\iff$ it is possible to derive this from the relation. The important part of the $\iff$ part is the forwards implication, the contrapositive of which is "if it cannot be shown, then it is not true". Unprovability is essentially what you need to prove! Unsurprisingly this is hard, hence the Novikov–Boone theorem.

The way to deal with these problems is to show that these relations hold $\iff$ (proving both ways) another canonical set do. Then the groups must be the same.

(As pointed out in your other question, you can deduce unprovability by explicitly constructing a model which violates the statement, though this in general is hard. For simple problems it should work.)

not all wrong
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  • Okay. I kind of got the idea that you say. So for example, as for $<i \mid i^n =1>$, this actually implicitly defines that $i^2 \neq 1, i^3 \neq 1$ and so forth, right? – le4m Jun 20 '13 at 08:22
  • I mean when the case is supposed to be $n >3$. – le4m Jun 20 '13 at 08:24
  • Anyway I think it is definite that we cannot derive $i^2 \neq 1$ nor $i^2 =1$ from $i^n=1$ where $n >2$. – le4m Jun 20 '13 at 08:33
  • So it is almost something like if we cannot derive $i^2=1$ from $i^n =1$ then we assume $i^2 \neq 1$. That is, if $i^2 =1$ is unprovable from $i^n =1$ then $i^2 \neq 1$. – le4m Jun 20 '13 at 08:38
  • Hm.. if it is indeed true, then for cyclic group, we can guess which is equal to which and which is not equal to which quite easily. But then as for the presentation for $Q$, it would be much more complex, isn't it? – le4m Jun 20 '13 at 08:40
  • It doesn't quite implicitly define $i^2\neq 1$ but it does imply this, because as you say you cannot derive this from the given relation. And yes, in general this is hard - see e.g. Novikov–Boone theorem. However, if you can show, for two sets of relations $R_1,R_2$ that $R_1\implies R_2,R_2\implies R_1$ then equality of groups follows. – not all wrong Jun 20 '13 at 09:11
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Assuming $i^2 = 1$ will give you the dihedral group which is distinct from the quaternion group. Call $s = i$ and $r = j$, which gives the group $\langle r,s \, | \, r^4 = s^2 = 1, sr = r^3 s \rangle$. Since the quaternion group and dihedral group are distinct, you cannot have $i^2 = 1$ in $Q_8$.

Hope that helps,

Patrick Da Silva
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  • Straightforward approach. In fact, if $iji^{-1}=j^3=j^{-1}$ and $i^2=1$ then we have $(ij)^2=1$ which is an essential relation in $D_8$. (+1) – Mikasa Jun 20 '13 at 01:22
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Hint. Usually you see this type of presentation with $j^3i$ written as $k$. We know that $ij=k$, so $i^2j=ik$. If $i^2=1$, then $ik=j$. But we also know from the standard presentation that $ik$ should be equal to $-j$. Can you use this as a guide to reach a contradiction with your form of the presentation?

Sequoia Grey
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  • What is $-j$? :P In the presentation $\langle i,j , | , i^4 = j^4 = 1, ij = j^3 i \rangle$, there is no mention of a minus sign. – Patrick Da Silva Jun 20 '13 at 01:46
  • @PatrickDaSilva I'm refering to the standard presentation of $Q_8$: $$\langle i,j,k,-1|i^4=j^4=k^4=(-1)^2=1,ij=k,jk=i,ki=j,ji=(-1)k:=-k,kj=(-1)i:=-i,ik=(-1)j:=-j\rangle$$ This is how you usually see it, and is easily Googlable. Again my answer was intended as a hint, that $j^3i$ corresponds to $k$ in the standard presentation, and to use that as a guide towards the contradiction proof with OP's presentation. – Sequoia Grey Jun 21 '13 at 17:55