Algebraic Structures that do not respect isomorphism 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!
 A: 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.
A: 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 $A,B$.
A: 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.
A: 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 $X / \operatorname{im} f$).
A: 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 $C$. 
