Algebraic Structures that do not respect isomorphism

Solution 1:

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.

Solution 2:

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 $X / \operatorname{im} f$).

Solution 3:

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.

Solution 4:

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 $A,B$.