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This is a pet peeve of mine. Both my textbook and my instructor, often say $G$ is $G'$, when really what they mean is $G$ is isomorphic to $G'$. Why is this an accepted convention?

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Usually people do this when there is a canonical choice of isomorphism. – Andrew Sep 28 '12 at 16:38
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It depends on context. It is very important to be able to be both pedantic and flexible about this depending on the situation. – Phira Sep 28 '12 at 16:40
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Often things are only defined up to isomorphism, for example anything defined using category theory. – Alex Becker Sep 28 '12 at 16:50
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This seems very strange to me. Under what circumstances would you consider it to be correct to say that a group "is" $\Bbb Z_6$? Maybe you want to say that only $\{0,1,\ldots 5\}$ with addition modulo 6 "is" $\Bbb Z_6$. But if there is a sense in which that object "is" $\Bbb Z_6$ and $\{NIL, I, II, III, IV, V\}$ (with suitable addition) "is not" $\Bbb Z_6$, I can't imagine what it would be. But if you can't identify a circumstance under which it would be correct to say that some group "is not" $\Bbb Z_6$ but is "isomorphic to" $\Bbb Z_6$, then it is a distinction without a difference. – MJD Sep 28 '12 at 17:02
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Actually saying they are equal becomes problematic if you try to compare elements from each group by, say, setting them equal to each other without the intervening isomorphism. This sort of confusion has happened to me when working with several layers of cosets - it's important to remember the isomorphism because without it you end up trying to say that some element is equal to some coset, which doesn't make sense. – Michael Boratko Sep 28 '12 at 20:19
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8 Answers

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It is the central insight of mathematics of structures that things that seem to be different are "really" the same thing. It doesn't help understanding if you are not able to switch the usage of "same" according to context. After all, each pair of objects is different, so what would 1+1 actually mean? And no, it is not a good idea to try to do mathematics as a formal game without any meaning.

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I like the comparison of "1 fire truck plus 1 fire truck is 2 fire trucks versus 1 apple plus 1 apple is 2 apples" to "Sym(3) acting on {1,2,3} has a unique non-identity normal subgroup consisting of the even permutations versus Sym(3) acting on {x,y,z} has a unique non-identity normal subgroup consisting of the substitutions leaving (x-y)(y-z)(z-x) invariant." S3 is the name of the groups that act a certain way. 1 is the name of the quantities that act a certain way. – Jack Schmidt Sep 28 '12 at 17:28
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Because the real object is only defined up to an isomorphism. The thing you think is $G$ is really just a model, as is $G'$.

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I disagree with your statement that the "real object" only being defined up to isomorphism. If you look in any(most) basic algebra books, the definition is never up to isomorphism. We start talking about isomorphic groups after the definition is given. – Thomas Sep 28 '12 at 17:08
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@Thomas I don't think you really understand what he is saying. – Makoto Kato Oct 2 '12 at 22:34
@MakotoKato: You are probably right. But I still don't think that the answer is good even is I actually do misunderstand him. Maybe he could clarify to help me understand? – Thomas Oct 2 '12 at 23:31
@Thomas you asked why people say the object is the same when in fact as point sets the objects are different. The reason is that the structure which defines the group is really the heart of the matter. There are many objects which can be given a particular structure. For a non-group example, $a+ib \in \mathbb{C}$ can be viewed as $\mathbb{R}^2$ with $(a,b)*(c,d) = (ac-bd,ad+bc)$. On the other hand, $a+ib \in \mathbb{C}$ can be viewed as $\left[ \begin{array}{cc} a & -b \\ b & a \end{array} \right]$. You can also view $\mathbb{C} = \mathbb{R}[x]/<x^2+1>$. So, which is $\mathbb{C}$? – James S. Cook Oct 3 '12 at 2:31
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@JamesS.Cook: Just to point out: I am not the OP. I, in fact, wrote my own answer to this question. I do still disagree with your statement that "...the real object is only defined up to an isomorphism." This is simply not true. Two isomorphic groups are indeed not the same thing. They are different. As I said in the first comment: If you look in any(most) basic algebra books, the definition is never up to isomorphism. – Thomas Oct 3 '12 at 2:54
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up vote 4 down vote

It is a question of language and of ontology (the essence of what a thing "is").

An Isomorphism can be seen as making a connection between two different aspects of the same abstract reality, thus illuminating its "true" nature more fully than either aspect.

Alternatively:

An Isomorphism can show us that two distinct entities with their separate realities look the same when seen from a particular perspective.

The concept of isomorphism does both jobs - hence the ambiguity - which is rarely a problem, unless you can give a specific example ... ?

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As an example of the second $\mathbb Z$ and $2 \mathbb Z$ are isomorphic as (additive) abelian groups - but that does not exhaust the interesting mathematical relationship between them. – Mark Bennet Sep 28 '12 at 17:33
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Isomorphism is just a renaming of elements.

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Something I find useful is to think about abstract groups and realizations of each of them.

You could say that an abstract group is just a collection of objects with particular relations holding among them, defined through the group product (whatever it may be). You can completely specify an abstract group by writing down the multiplication table or its presentation (not an easy task, generally speaking).
Following this approach, you can say that

  • there is only one group of three elements: its elements are $e,a,b$, where $e\cdot a = a\cdot e = a$, $e\cdot b = b\cdot e = b$, $a\cdot a = b$, $a\cdot b=b\cdot a = e$, $b\cdot b = a$ (sorry if I didn't write nicely the multiplication table). Then its presentation is $\langle a\;|\;a^3=e\rangle$ and you call it "cyclic group".
  • there are two groups of four elements: one has the presentation $\langle a\;|\;a^4=e\rangle$ (cyclic group); the other is given by $\langle (a,b)\;|\;a^2=b^2=(ab)^2=e\rangle$ (dihedral group).

Now observe that you can interpret the group of order 3, usually denoted by $\mathbb{Z}_3$, as the set $\mathbb{Z}/3\mathbb{Z}$ of residue classes with addition$\mod 3$, or the set of the complex cubic roots of unity with the ordinary product over $\mathbb{C}$.
Similarly, the cyclic group $\mathbb{Z}_4$ is the group $\mathbb{Z}/4\mathbb{Z}$ of residue classes$\mod 4$, or the group of fourth roots of unity. The dihedral group $D_2$ is usually referred to as the group of simmetries of a segment (i.e. a degenerate regular polygon with two sides).

All of these groups, then, admit different realizations in terms of matrices: these are usually called group representations (e.g. you make use of rotation and reflection matrices to construct the 2 dimensional representation of every dihedral group). The sets of matrices used to build a representation are subgroups of $\text{GL}(n)$.

Every realization of a group is isomorphic to all the others. Hope this helps!

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up vote 1 down vote

Ask yourself what a group really is. It is a set together with an operation/composition that satisfies certain relations. In many ways it is the composition is what determines all the properties of the group. If you for example ask about the number of subgroups of the group or whether or not there are any normal subgroup or if the group is abelian, then it all depends on the composition. If you change the underlying set by just calling the elements something else, the all these properties remain the same. This is all the same when we for example talk about the representation theory of a group. The number of irreducible representations of a group remain the same.

However, the definition of a group is not a "model" (as in another answer). I think that it is wrong to say that a group is always defined up to isomorphic. If that was so, then we wouldn't need to talk about groups being isomorphic ...

So in the end we can think of two isomorphic groups as being the same group even though they strictly speaking are not.

Can one make this formal? Yes, one can simply talk about the set of groups and then define an equivalence relation where $G \sim H$ if they are isomorphic. Then you can "mod out" by this equivalence relation. So you get a set of equivalence classes. This would justify talking about isomorphic groups as being the same.

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Clearly, they are not the same unless the whole system is adjusted so that they are made to be the same. Okay, what I mean by this is, for example, that when you construct $\mathbb{R}$ from $\mathbb{Q}$, you are actually constructing a new object $\mathbb{R}$ that does not at all include $\mathbb{Q}$ as a subset. But then the thing is that this new constructed object $\mathbb{R}$ contains a subset which behaves exactly in same the way as $\mathbb{Q}$ does, that is, isomorphic. So in this case, we wholly discard the previous $\mathbb{Q}$ but then adopting the whole new system $\mathbb{R}$ we redefine that subset as $\mathbb{Q}$ (using the same symbol). Note that though the previous rational number system and the new rational system are denoted by the same symbol, they are never the same logically.

Okay back to actual point. The reason why we simply say 'is' or consider them as 'equal' is only for convenience, the convenience to avoid all the detailed formalisms. If you need to deal with those groups in the fully formalized rigorous context, you should not make them as the same. But often times, you can prove just 'intuitively', and since to make them into a fully formalized version is a mere labor (though it maybe very long time-consumming labor), you just avoid them.

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up vote 0 down vote

It depends on context. Sometimes it's ok to say they are the same and sometimes it isn't.

An isomorphism is just a renaming of elements, so as long as the names don't actually mean anything or are unconnected to anything then it's ok. Structural information about a group such as how many subgroups it has doesn't depend on the names of the elements.

But as soon as the names of the elements are related to something else then the labelling of elements does matter. See for example the question about whether two rings can be non-isomorphic to each other even though the operations from one ring are isomorphic to the operations in the other ring.

Also the question "How many automorphisms does a group have?" requires you to both distinguish between relabellings and equate them (as isomorphic) at the same time.

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