Algebraic Structures that do not respect isomorphism

Question

One of the first things a student learn in Algebra is isomorphism, and it seems many objects in algebra are defined up to isomorphism.

It then comes as a mild shock (at least to me) that quotient groups do not respect isomorphism, in the sense that if $G$ is a group, and $H$ and $K$ are isomorphic normal subgroups, $G/H$ and $G/K$ may not be isomorphic. (see Isomorphic quotient groups)

My two questions are:

1) What other algebraic "structures" or "operations" do not respect isomorphism?

2) Philosophical (or heuristically), why are there algebraic structures that do not respect isomorphism? Is this supposed to be surprising or not surprising? To me $G/H$ not isomorphic to $G/K$, even though I understand the counterexample, is as surprising as $\frac{2}{1/2}\neq\frac{2}{0.5}$.

Thanks for any help!

On 2): not surprising to me. The statement that two distinct normal subgroups are isomorphic leaves open how these subgroups are related with the original group. — drhab, Jul 03 '16 at 15:26
If you have an automorphism of $G$ which sends $H$ to $K$, then the quotients are isomorphic. Therefore I would agree with drhab: The reason is that the isomorphism between the normal subgroups doesn't say anything about how these subgroups relate to the rest of the group. — , Jul 03 '16 at 15:51
@StefanPerko No one seems to have mentioned natural isomorphisms yet. :) — almagest, Jul 03 '16 at 15:52
A simpler example with no algebra, just plain sets. Let $A$ be a set, and let $B,C$ be subsets of $A.$ Amazingly, knowing that $B$ and $C$ are isomorphic (i.e. have the same cardinality) does not guarantee that $A\setminus B$ and $A\setminus C$ are isomorphic!! — bof, Oct 10 '16 at 09:11
@bof If A is finite then they surely $A \setminus B$ and $A\setminus C$ have the same cardinality. I don't see how they could not, even if A is infinite. — RyeCatcher, Feb 03 '23 at 06:13
@RyeCatcher What if $A={1,2,3,\dots}$ (the set of all positive integers), $B={3,4,5,\dots}$, and $C={6,7,8,\dots}$? — bof, Feb 04 '23 at 04:34

score 34 · Answer 1 · 2016-07-03T19:30:53.587

The problem is that you have the wrong notion of isomorphism!

It's not enough that $H$ and $K$ be isomorphic as groups — what we want is for the embeddings $H \to G$ and $K \to G$ to be isomorphic as diagrams of groups: in this case we need a commutative diagram

$$ \require{AMScd} \begin{CD} H & @>1>> G \\ @VV \varphi V @VV \theta V \\ K & @>1>> G \end{CD} $$

where $\varphi$ and $\theta$ are isomorphisms. If this is true, then $\bar{\theta} : G/H \to G/K$ will be a well-defined isomorphism.

More generally, there is a notion of homomorphism between such diagrams: if we have two group homomorphisms $A \xrightarrow{f} B$ and $C \xrightarrow{g} D$, then a homomorphism from the former to the latter is a commutative square

$$ \require{AMScd} \begin{CD} A & @>f>> B \\ @VV \varphi V @VV \theta V \\ C & @>g>> D \end{CD} $$

that is, a homomorphism from $f$ to $g$ is a pair $(\varphi, \theta)$ of group homomorphisms with the property that $\theta \circ f = g \circ \varphi$.

Such homomorphisms compose in the obvious way, and the identity homomorphism is the one where $\varphi$ and $\theta$ are identity maps. An isomorphism is an invertible homomorphism, which in this case means that the two component group homomorphisms are invertible.

I like this answer a lot. It makes me wonder if it wouldn't be better to just stop emphasizing objects so much. So instead of $A/B$ the notation is $\mathrm{coker}(f)$ (for cokernel of $f$), where $f$ is the inclusion $A \hookrightarrow B$. Of course, the morphism needn't be injective, as your answer rightly emphasizes. — goblin GONE, Jul 04 '16 at 15:24
@goblin: In the case of vector spaces, you can take matrix algebra as a prototype for doing so! When working with just the standard vector spaces ($\mathbb{R}^n$), the objects themselves are almost universally ignored to the point where you might as well just use natural number labels for them. (e.g. one doesn't even bother to consider vectors in $\mathbb{R}^n$; instead one works with elements of $\hom(\mathbb{R}, \mathbb{R}^n)$ represented as $n \times 1$ matrices) — , Jul 04 '16 at 18:12

score 30 · Accepted Answer · edited Oct 10 '16 at 11:41

For a very simple example of this: $2\mathbb{Z}$ and $3\mathbb{Z}$ are isomorphic as abelian groups, but $\mathbb{Z}/2\mathbb{Z}$ is not isomorphic to $\mathbb{Z}/3\mathbb{Z}$.

Okay, so what gives?

The relevant concept here is that of a slice category. The point really is that $2\mathbb{Z}$ and $3\mathbb{Z}$ aren't just objects of $\mathbf{Ab}$, they're actually objects of the slice category $\mathbf{Ab}/\mathbb{Z}$ (the $/$ doesn't mean a quotient, it means a comma category.) Viewed as objects of $\mathbf{Ab}/\mathbb{Z}$, the objects $2\mathbb{Z}$ and $3\mathbb{Z}$ aren't isomorphic.

The lesson, really, is that if we're just given abelian groups $Y$ and $X$, the quotient $X/Y$ isn't automatically well-defined; we need to have a distinguished way of viewing $Y$ as an object of $\mathbf{Ab}/X$. In other words, we need a distinguished morphism $f:Y \rightarrow X$. It's probably best to assume $f$ is injective, too, but strictly speaking, this isn't really necessary (if $f$ is not injective then the result is the cokernel of $f$, roughly speaking the quotient $X / \operatorname{im} f$).

Just a small remark, in your last paragraph you could include something about "cokernels" I think. — Najib Idrissi, Oct 10 '16 at 09:06
@NajibIdrissi, I've made the answer CW; feel free to change it as desired. (I'm a little short on time right now.) — goblin GONE, Oct 10 '16 at 09:08
Done (but it wasn't a small change, you didn't have to make it CW, you probably deserve the rep...) — Najib Idrissi, Oct 10 '16 at 11:41

score 9 · Answer 3 · edited Apr 13 '17 at 12:20

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1) see this for some non-examples

2) As drhab put it:

"The statement that two distinct normal subgroups are isomorphic leaves open how these subgroups are related with the original group."

A normal subgroup $H$ is not "just" a certain subset of $G$, but it always comes with a monomorphism $H\to G$, the inclusion. You can think of this morphism as the "actual" subgroup. Two monos $h : H \to G$ and $k : K \to G$ are isomorphic, if there is a group isomorphism $i : H \to K$, such that (!) $k\circ i = h$. If two subgroups are isomorphic in this fashion, then the quotients $G/H$ and $G/K$ should be too.

edited Apr 13 '17 at 12:20

Community

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answered Jul 03 '16 at 15:32

Stefan Perko

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See goblin's answer for the categorical notion, that describes this idea of "isomorphism of subobjects". – Stefan Perko Jul 03 '16 at 15:52
$H := \mathbb{Z}/2\mathbb{Z} \times {0}, {0} \times \mathbb{Z}/2\mathbb{Z} \subseteq \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} =: G$ should show that your last assertions are wrong. – Jul 03 '16 at 15:57
There is some error here. Maybe it was supposed to say $i:G \to G$. – Max Jul 03 '16 at 19:22
@Max Thank you. Clearly I need to think this over again on another day ....... sigh – Stefan Perko Jul 03 '16 at 19:46
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We all have days like this. No worries :) – Max Jul 03 '16 at 19:59
See also Hurkyl's answer for the diagram. – filipos Jul 04 '16 at 12:29

user21820 · Answer 4 · 2016-07-04T12:57:41.810

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I know there are already excellent answers that deal with the mathematical question here about quotient groups, but I would like to address the part of the question about 'intuition'.

Philosophical (or heuristically), why are there algebraic structures that do not respect isomorphism?

If you want a structure-preserving map from $T$ to $U$, then you need it to preserve all the structure. If $T = f(S,A)$ and $U = f(S,B)$, then clearly it's not enough to have $A,B$ isomorphic, since that only guarantees that the size and internal structure of $A,B$ are the same, and whatever additional structure that $f$ imbues its output can be completely 'unknown' to $A,B$.

edited Jul 04 '16 at 12:57

answered Jul 04 '16 at 00:56

user21820

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How is $(\Bbb Z,<)$ isomorphic to $(\Bbb Q,<)$ as a total order? One is dense and the other isn't. – Mario Carneiro Jul 04 '16 at 08:48
@MarioCarneiro: Lol you're of course right. What was I thinking?! Is there a good example to replace that? I didn't want a trivial kind of isomorphism where I just discard all the operations of the ambient structure. – user21820 Jul 04 '16 at 12:54
In the same vein, $(\Bbb Q,<)$ is isomorphic to $(\Bbb Q^+,<)$ as a linear order, but they embed differently inside $(\Bbb R,<)$: one has a lower bound and the other doesn't. Not sure what kinds of derived structure you could build on that... – Mario Carneiro Jul 04 '16 at 13:03
@MarioCarneiro: Yea I thought about that example, but as you say I don't a natural derived structure that depends on that distinction. Sigh this isn't the only sleepy mistake I made today. =( – user21820 Jul 04 '16 at 13:35
Well, if you want to make a quotient total order $A/S$, you could let $x\sim y$ if there exist $s\le x\le y\le s'$ for $s,s'\in S$ or vice-versa. Under this equivalence, $\Bbb R/\Bbb Q$ and $\Bbb R/\Bbb Q^+$ are equivalent as partial orders to ${0}$ and $(-\infty,0]\cup{1}$, respectively. – Mario Carneiro Jul 05 '16 at 01:46

score 2 · Answer 5 · answered Oct 10 '16 at 09:03

Perhaps this is the Galois-theoretic analogue for fields of the example for groups. $A={\bf Q}(\root4\of2)$ and $B={\bf Q}(i\root4\of2)$ are isomorphic as fields, but not as extensions of $C={\bf Q}(\sqrt2)$. That is, there is no field isomorphism of $A$ and $B$ fixing $C$.

Algebraic Structures that do not respect isomorphism

5 Answers5