Confusion regarding the identity (or non-identity) of two groups

Question

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?

Answer 1

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.

Answer 2

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.