Recently I was perplexed by a simple abstract algebra question. The question was as follows,

We know that $D_4$ (the Dihedral Group of order $8$) and $Q_8$ (the Quaternion Group) are the only non-abelian groups of order $8$ upto isomorphism. Now does this mean that if $G$ be any non-abelian group of order $8$ then either $G=D_4$ or $G=Q_8$?

The immediate answer is that it is not the case (provided we assume that usual definition of "$=$"). More precisely, if $G$ be a given non-abelian group of order $8$ then $G$ may not be identical either to $D_4$ or to $Q_8$ elementwise, but $G$ will be isomorphic to exactly one of them.

Now let us come to the presentation of $Q_8$. It is (as our professor told us), $$\langle a,b\mid a^4=1,a^2=b^2,b^{-1}ab=a^{-1}\rangle$$Then he said that,

If we replace $a$ by $x$ and $b$ by $y$ then the group thus obtained will be isomorphic to $Q_8$ and not identical to $Q_8$.

This is the part where I am a bit confused. My question is,

Aren't $a,b,x,y$ simply some symbols which we use to denote the elements of a group? If so then why the groups are not identical (assuming the operations on the groups to be identical) even if we denote the same element by different symbols? It is true that the groups are not syntactically identical but are they not semantically identical?

  • 1
    $\begingroup$ The way I see it, the groups can be different groups but isomorphic to each other. For example, the group of symmetries of a triangle (dihedral group of order 6) and the permutation group $S_3$ are isomorphic, but the way we interpret elements are slightly different. I guess the groups are different in a sense, though they're isomorphic. $\endgroup$ – Camille Feb 4 '17 at 3:44
  • $\begingroup$ @Camille: We assume the operations to be identical (please read the last highlighted box again). $\endgroup$ – user170039 Feb 4 '17 at 3:49
  • $\begingroup$ @user170039 I guess you can't have identical operations where the sets are not identical, because an operation is a subset of $S \times S \times S$. $\endgroup$ – Eric Auld Feb 4 '17 at 3:54
  • $\begingroup$ @EricAuld: Yes that's what has been confusing to me and one of the main reason to post the question here. $\endgroup$ – user170039 Feb 4 '17 at 4:01
  • $\begingroup$ Well, then let's say, $\mathbb{Z}$ and $2\mathbb{Z}$. They're both groups, they're isomorphic, and the operation is addition. But they're not exactly the same set, yes? $\endgroup$ – Camille Feb 4 '17 at 4:34

Formally, $\langle a,b\mid a^4=1,a^2=b^2,b^{-1}ab=a^{-1}\rangle$ is defined as the free group in alphabet $a,b$, divided by the normal subgroup generated by $a^4$, $a^2b^{-2}$, and $ab^{-1}ab$. Free group in turn is defined as the set of equivalence classes of finite sequences... etc. etc.

If you start with a different (disjoint) alphabet, you will simply obtain a disjoint set, as far as set theory is concerned. So in terms of sets, $a,b$ are not just symbols we use to denote the elements of the group, but rather the symbols we use to construct the group.

On the other hand, from more structural point of view, asking whether an arbitrary group $G$ is equal to $D_8$ or $Q_8$ is simply nonsense, as there is no notion of equality between two distinct groups, unless they are both contained in a larger structure. You could say that this larger structure is the set-theoretic universe (indeed, that is the point of view taken in the preceding paragraphs), but in vast majority of cases, this kind of equality is not of much interest, unlike, say, equality of two subgroups of a third group.

| cite | improve this answer | |
  • $\begingroup$ What are $a,b$? Are they not variables of the language of Formal Group Theory, say following the axioms as written here? $\endgroup$ – user170039 Feb 4 '17 at 4:10
  • $\begingroup$ Furthermore under the assumption that the operations on the groups are identical doesn't it follow that $(a,b)=(x,y)$? $\endgroup$ – user170039 Feb 4 '17 at 4:13
  • $\begingroup$ They are elements of the alphabet upon which we are building the free group. To be even more concrete we could say $\{0,1\}$ versus $\{2,3\}$. Element of the free group generated from $\{0,1\}$ are sequences like $1001^{-1}0110^{-1}1011$ while those of the free group generated from $\{2,3\}$ are sequences like $232^{-1}23323^{-1}223$. $1001^{-1}0110^{-1}1011$ is simply not an element of the free group generated by $\{2,3\}$. $\endgroup$ – Derek Elkins left SE Feb 4 '17 at 6:42
  • $\begingroup$ @user170039: What do you mean by "operations are identical"? Under the usual construction, no, they are not variables, but rather some specific (distinct) elements. For example, they can be the letter $a$ and the letter $b$. $\endgroup$ – tomasz Feb 4 '17 at 11:01

When you talk of a group in terms of presentations (that is generators and relations) we make no mention of the binary operations: or rather assume the binary operation is concatenation of symbols. Under this implicitly assumed binary operation then we may accept that it is simply a different name for the symbol.

But in reality things could be different: For example consider, for each integer $n\ge3$, the set $G_n=\{1, n-1\}$. I make each of this set into a group by the binary operation of multiplication mod $n$. All of them are cyclic groups of order 2. So they are all different avatars of the same group, but not the same.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy