From what I understand about isomorphisms is that two isomorphic groups are the same groups. They may have different names for the same elements and the operation. But the point is that the groups are same since their elements combine the same way.

So if $G \cong G'$ then every group property about $G$ also holds for $G'$, am I correct?

Now my question is— Can I interchange $G$ and $G'$ whenever and wherever I want? I thought the answer was obviously yes but... now I'm not sure.

For example: $\mathbb{Z} \cong \mathbb{2Z}$ so shouldn't $\mathbb{Z}/\mathbb{Z} = \mathbb{Z}/\mathbb{2Z}$ under the group operation $(a+H) + (b +H) = (a+b) + H$? where $H$ is either $\mathbb{Z}$ or $\mathbb{2Z}$ since both are the same groups.

But that's clearly not the case as $\mathbb{Z}/\mathbb{Z} = \{0\}$ but $\mathbb{Z}/\mathbb{2Z}=\{0,1\}$ So where does one draw the line between two isomorphic groups? How "same" are two isomorphic groups? I'm very confused.

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    $\begingroup$ It is true that $\mathbb{Z} \cong \mathbb{2Z}$. However, the issue for the quotient groups here is that $2 \mathbb Z \subsetneq \mathbb Z$. $\endgroup$ Dec 27, 2021 at 16:42
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    $\begingroup$ The freedom to rename is not a freedom to be inconsistent. If you want to rename the elements in $\mathbb Z$ so that $1$ is renamed to $2$, $2$ to $4$, $3$ to $6$ etc. - then of course the elements of $2\mathbb Z\subseteq \mathbb Z$ must be renamed using the same procedure. You will end up with $4\mathbb Z\subseteq 2\mathbb Z$, and indeed $\mathbb Z/2\mathbb Z\cong 2\mathbb Z/4\mathbb Z$. $\endgroup$
    – user700480
    Dec 27, 2021 at 16:43
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    $\begingroup$ Why is this closed under duplicate? I'm asking about a particular example and the difficulty I'm facing with it. How is the question same as the other? I'd already read that one. It's where I came from. The accepted answers says exactly the same thing as my first paragraph. But my question is different. $\endgroup$
    – William
    Dec 27, 2021 at 16:46
  • $\begingroup$ Presentations of the same group (up to isomorphism) can have different properties. $\endgroup$
    – Shaun
    Dec 27, 2021 at 16:54
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    $\begingroup$ Frankly, I regard the "s" word as a 4-letter word in mathematics. I would suggest that you learn to love the 10-letter "i" word in your title. And then if you can extend that to loving the 11-letter word "isomorphism", and to think about actual isomorphisms as useful and interesting mathematical objects themselves, even better! $\endgroup$
    – Lee Mosher
    Dec 27, 2021 at 18:51

6 Answers 6


This is a great question and your example with group quotients perfectly highlights why one has to be careful about notions of identity when doing maths. To answer your question,

Can I interchange G and G′ whenever and wherever I want?

The answer is yes, as long as you are 100% sure that you are thinking of both groups as literally just groups and nothing more (people will sometimes use the phase 'up to isomorphism' to convey this idea). This resolves the problem you had with quotients: you can't quotient an arbitrary group by another arbitrary group, the group by which you are quotienting has to be a subgroup of the other. This is extra 'data' which can be encoded in multiple ways, one nice way is to consider the special injection that embeds the one group inside the other (which doesn't exist for two arbitrary groups). So to recap, subgroups aren't just groups, they're groups with some extra data and $2\mathbb{Z}$ and $\mathbb{Z}$ are different subgroups of $\mathbb{Z}$ despite being 'the same' groups (i.e. isomorphic).

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    $\begingroup$ "the group by which you are quotienting has to be a subgroup of the other" but $Z$ and $2Z$ are both subgroups of $Z$ no? and I don't get the part where you say they are the same groups but different subgroups? I'm sure I'm missing something here but I don't get it. What is the extra data about subgroup? $\endgroup$
    – William
    Dec 27, 2021 at 17:14
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    $\begingroup$ Oh you mean they're different subgroups because they are made up of different subsets of the group? Oh okay. Fair enough. But what exactly go wrong here in my example? I'm not able to "see" it still :( $\endgroup$
    – William
    Dec 27, 2021 at 17:26
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    $\begingroup$ The problem in your example is that, to conclude that the quotient is the same, you have to start with 'the same' subgroups, but you didn't, you started with two different subgroups (that just so happen to be 'the same' as groups). $\endgroup$
    – Arthur
    Dec 27, 2021 at 17:28
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    $\begingroup$ So in total, there are four descriptions of mathematical objects: the group $\mathbb{Z}$, the group $\mathbb{2Z}$, the subgroup $\mathbb{Z}\leq\mathbb{Z}$ and the subgroup $\mathbb{2Z}\leq\mathbb{Z}$. The first two descriptions define the same mathematical object, the second two don't. $\endgroup$
    – Arthur
    Dec 27, 2021 at 17:46
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    $\begingroup$ Oh! Oh! Oh!! I get it now. If $H,H' ≤G$ and $H \cong H'$ then as groups, $H$ and $H'$ are same. But "subgroup" is also a relation and not an independent concept. So you always have subgroups of some group. So as subgroups of $G$, $H$ and $H'$ are different. So $G/H ≠ G/H'$ because quotient groups aren't just about $H$ but rather $H$ wrt $G$. If it were about only $H$ then one can interchange it with $H'$ is what you're saying. $\endgroup$
    – William
    Dec 27, 2021 at 18:12

Here is a question that is similar: when can replace equivalent things with one another?

So if we remember that an isomorphism of groups is a function $\phi:G\rightarrow G'$. With the example you gave you have $\phi:\mathbb{Z}\rightarrow 2\mathbb{Z}$, then your issue was that $\mathbb{Z}/\mathbb{Z}\not\cong\mathbb{Z}/2\mathbb{Z}$. However, the issue is where are you applying the homomorphism. You are saying that $G/G\not\cong G/\phi(G)$; however, we have that $\phi(G)\subset G'$ and $\phi(G)\not\subset G$ (although it may be isomorphic to some subgroup of $G$ like in your example).

The key use of an isomorphism is that you can take a problem in one setting and view it in another setting through the isomorphism. Thus, what would be true is that if $\phi:G\rightarrow G'$ is an isomorphism, then if $H\lhd G$, then it will be the case that $G/H\cong \phi(G)/\phi(H)$.

I think the long story short is that when you have an isomorphism between two objects you must think of them as living in different spaces. If you are familiar with the lattice of subgroups I think this is a nice way to view it, since if you have $\phi: G\rightarrow G'$ where $G=\mathbb{Z}$ and $G'=2\mathbb{Z}$ since these are isomorphic groups their lattice of subgroups are isomorphic and we should think of $\mathbb{Z}$ to be the maximal element in the lattice of $G$ and $2\mathbb{Z}$ to be the maximal element in the lattice of $G'$. Even though we have that $2\mathbb{Z}$ appears in the lattice of $G$, the corresponding element in $G'$ would be the subgroup $4\mathbb{Z}$, so under the isomorphism it would not make sense to "replace $2\mathbb{Z}$ in the $G$ world with $2\mathbb{Z}$ in the $G'$ world" as they aren't the same under the isomorphism. What we would do is replace the $2\mathbb{Z}$ with the corresponding subgroup in the isomorphism namely $4\mathbb{Z}$, and when we do so we do have that $\mathbb{Z}/2\mathbb{Z}\cong 2\mathbb{Z}/4\mathbb{Z}$.


There is a precise way to formalise this.

We want to speak of sets not as collections of particular mathematical objects but as a bag of featureless dots. These bags of dots are related to each other by functions, which also relate dots in one bag to dots in another.

In order to do this, we propose a language for discussing sets. Capital letters will represent sets, and lowercase variables will represent functions between sets.

We define a "nice proposition about sets" as any proposition that can be generated from the following rules:

  • For any sets $A, B$ and any function variables $f, g: A \to B$, $f = g$ is a nice proposition
  • We can combine nice propositions $\phi, \psi$ using the operators $\land$, $\lor$, $\neg$, and $\implies$ to get another nice proposition
  • $\top$ and $\bot$ (that is, true and false) are nice propositions
  • If $\phi(A)$ is a nice proposition, where $A$ is a set variable, then $\forall A (\phi(A))$ is a nice proposition (and similarly for $\exists$)
  • For all set variables $A, B$, if $\phi(f)$ is a nice proposition (where $f : A \to B$ is a function variable), then $\forall f : A \to B (\phi(f))$ is a nice proposition (and similarly for $\exists$)

Finally, in our language, we can form new functions from old ones using function composition and also discuss identity functions. Note that we will only discuss function composition when we know that the functions are composable syntactically.*

Note that we very deliberately avoided two things. First, there is no mention of the elementhood relation at all. Second, there is no way to compare two sets for equality, nor is there a way to compare two functions for equality unless the two have the same domain and codomain. This is deliberate. Also note that all variables here have specific types, and that we rely on these types to determine whether we can discuss equality.

Despite this avoidance, it is both an empirical fact and a (rather complicated) theorem that all propositions with no free variables can be translated to a "nice proposition about sets". For most mathematically useful propositions, the translation is relatively intuitive for those comfortable with category theory.

To get around the restriction that we not discuss elements directly, we can fix some 1-element set $1$ and discuss elements of $A$ as functions $1 \to A$. We denote the situation $x : 1 \to A$ as $x :\in A$. When we have $x :\in A$ and $f : A \to B$, we write $f \circ x$ as $f(x)$ (note that $f(x) :\in B$). Note that this is enough to define, for example, a group (as a set $G$, together with a function $- \cdot - : G^2 \to G$ which satisfies the group laws**).

We thus have the following theorem:

Big Theorem. Consider some nice proposition $\phi(G)$ - that is, some statement $\phi$ where the group variable $G$ occurs free. Then we can prove the following statement: "For all groups $G_1$, $G_2$, if $G_1$ and $G_2$ are isomorphic and $\phi(G_1)$, then $\phi(G_2)$."

Let us discuss how this relates to your example of $\mathbb{Z}$ and $2 \mathbb{Z}$.

We may try to define $\phi(G)$ as the statement "$\mathbb{Z} / G$ is the zero group". Then taking $G_1 = \mathbb{Z}$ and $G_2 = 2 \mathbb{Z}$, we see that $\phi(G_1)$ holds but $\phi(G_2)$ is false. This would appear to violate our Big Theorem.

To understand what's going on, we need to understand exactly what is going on with the statement $\phi(G)$. Note that in this statement, it is necessary for $G$ to be a subgroup of $\mathbb{Z}$ - in particular, it must be a subset. That is, we must have $\forall x \in G (x \in \mathbb{Z})$.

Of course, we cannot express such a statement as a nice proposition about sets, since it relies on the proposition $x \in \mathbb{Z}$, which is not a nice proposition. So we have to approach things a bit differently.

Rather than forcing $G$ to be a subgroup in the traditional, literal sense, we instead make $G$ a subgroup of $\mathbb{Z}$ in a more general sense. That is, we discuss a group $G$, together with some injective group homomorphism $i : G \to \mathbb{Z}$. In other words, we discuss $G$ together with a specific way that $G$ is a subgroup of $\mathbb{Z}$.

This makes it clear that $\phi(G)$ is not really just a proposition about $G$ - it's also about the way that $G$ is a subgroup of $\mathbb{Z}$. So it's really a proposition $\phi(G, i : G \to \mathbb{Z})$. Because $\phi$ depends on a secondary variable $i$, we see that we cannot apply our Big Theorem to conclude that $\phi(\mathbb{Z}) \iff \phi(2 \mathbb{Z})$.

If it were the case that the isomorphism $w : \mathbb{Z} \to 2 \mathbb{Z}$ "played nicely" with the inclusion maps $i_1 : \mathbb{Z} \to \mathbb{Z}$, $i_2 : 2 \mathbb{Z} \to \mathbb{Z}$ (that is, if $i_1 = i_2 \circ w$), then we would be able to conclude that $\phi(\mathbb{Z}, i_1) \iff \phi(2 \mathbb{Z}, i_2)$ using a generalisation of our Big Theorem. But of course the isomorphism does not play nicely with the inclusion maps.

*An astute observer will note that this is exactly the language of category theory. A "nice proposition about sets" is just some statement about the category of sets (which avoids discussing equality of objects).

**Technically, we need a way to define $G^2$ first. This is done using the categorical definition of the universal property of the product.

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    $\begingroup$ Why is this answer being downvoted? $\endgroup$
    – William
    Dec 27, 2021 at 18:31

This is just an elaboration of Arthur's answer elsewhere in this thread. I thought you might like to see a specific example.

Consider $$G = \color{maroon}{\Bbb Z_2}\times \color{darkblue}{\Bbb Z_4}.$$

$G$ has a subgroup $A$ that is generated by $\langle 1, 0\rangle$:

$$\def\elt#1#2{\langle{#1},{#2}\rangle}A = \{\elt10, \elt 00\}$$

and another subgroup $B$ that is generated by $\langle 0, 2\rangle$:

$$B=\{\elt00, \elt 02\}$$

Both $A$ and $B$ are isomorphic to $\Bbb Z_2$, and so to each other. But the quotients $G/A$ and $G/B$ are not isomorphic. $G/A$ is like taking $G$, ignoring the first component, and keeping the second component intact. The result is $$G/A \simeq \color{maroon}{\Bbb Z_1}\times\color{darkblue}{\Bbb Z_4}.$$ $G/B$ is like taking $G$ and keeping the first component but ignoring everything but the parity of the second component. The result is $$G/B \simeq \color{maroon}{\Bbb Z_2}\times\color{darkblue}{\Bbb Z_2}.$$ Even though $A$ and $B$ are isomorphic, $G/A$ and $G/B$ are not isomorphic. (In particular, $G/A$ is cyclic and $G/B$ isn't.)

Each component of $G = \color{maroon}{\Bbb Z_2}\times \color{darkblue}{\Bbb Z_4}$ contains a factor of $\Bbb Z_2$. Each of $A$ and $B$ is isomorphic to $\Bbb Z_2$. When we construct the quotient of $G$ by $A$ or $B$, we get a different result depending on which component we divide the factor from.

Two groups that are isomorphic share the same internal structure. But a quotient $G/N$ doesn't depend only on the internal structures of $G$ and $N$. The quotient also depends on the relationship between $G$ and $N$. So even though $A$ and $B$ have the same internal structure, $G/A$ and $G/B$ are different, because those structures reside inside of $G$ in different ways.


Topological and illustrated version of Arthur's nice answer:

A trefoil knot $K_3$ in $\Bbb R^3$ cannot be unknotted but it can be in $\Bbb R^4$. Topologically a trefoil knot $K_3$ and a circle are homeomorphic i.e. one can deform one to other without cutting or gluing. So these two (one in $\Bbb R^3$ and one in $\Bbb R^4$) are the same but one uncovers a branch of mathematics known as Knot_theory while other one is fruitless from this point of view and application.

enter image description here

I think theses type of "same things" is a categorical notion but I am not sure.

same is true for $y=x$ and $y=\exp(x)$; both topologically are homeomorphic to $\Bbb R$, but one contains only positive numbers. One can say similar properties by considering them as additive and multiplicative groups.

These might be of interest to you:


Let me for convenience, first, assume all subgroups are normal, or that we are in an abelian group.

We form quotients of a group by its SUBgroups, not by another abstract GROUP. Subgroup means a SPECIFIC subset (which is closed ....) As groups Z and 2Z are "same". But as subsets of Z they are different.

Now to non-abelian case. Take $G= \{+1,-1\}\times S_3$ where the first factor $H$ is a group (of order 2) wrt multiplication of numbers.

This group $G$ has more subgroups of order 2 (isomorphic to H as a group) namely ones generated by a transposition in the second factor $S_3$. These subgroups are NOT even normal in $G$. SO quotient does not even exist by these subgroups.


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