When can we "switch" isomorphic things

Question

I'm finishing a course on basic abstract algebra, which covers groups, rings, modules, and finite group representations. However, up to this point, I am not very sure about the concept of isomorphism.

I know there are examples where given a group $G$ and two normal subgroups, $H$ and $K$, where $H \cong K$, but $\frac{G}{H} \not \cong \frac{G}{K}$. For instance, see here. However, for (external) direct sum, this does not seem to be a problem, for groups $A,B,C,D$, if $A \cong B$, $C \cong D$, $A \oplus B \cong C \oplus D$.

I only know a little (if not nothing) about category theory. When learning general topology, we constructed product topology and quotient topology with universal product, and that's the most I know about these constructions.

So my question is, when we create a new object based on two or more objects in the same category, does the isomorphism carry over? For instance if we make object $(AB)$ from $A$ and $B$, and $(CD)$ from $C$ and $D$, when can we conclude that $A \cong C$, $B \cong D$ implies that $(AB) \cong (CD)$, or vice versa?

Answer 1

This is answered by the concept of a functor. Notice that every functor $F : \mathcal{C} \to \mathcal{D}$ has the fundamental property $$A \cong A' \implies F(A) \cong F(A')$$ for all objects $A,A'$ in $\mathcal{C}$. Whenever you encounter a statement such as "if ... are isomorphic, then ... are isomorphic", it is usually a consequenc of this general result.

For example, the direct sum of abelian groups is a functor $$F : \mathbf{Ab} \times \mathbf{Ab} \to \mathbf{Ab},~ (A,B) \mapsto A \oplus B,$$ with the evident definition on morphisms. It follows that if $(A,B) \cong (C,D)$ (which is clearly equivalent to $A \cong C \wedge B \cong D$), then $A \oplus B \cong C \oplus D$.

The counterexample with quotient groups that you mentioned happens essentially because $H \mapsto G/H$ is not a functor on the category of groups that happen to be normal subgroups of $G$. It is a functor when you restrict the morphisms to inclusions, since any inclusion $H \subseteq H'$ induces a homomorphism $G/H \to G/H'$. So $G/(-)$ is a functor in this sense, but this only gives us $H = H' \implies G/H \cong G/H'$ what is not very interesting.

More interesting is the following category $\mathbf{Grp}_{\mathrm{sub}}$. Its objects are pairs $(H,G)$ where $G$ is a group and $H$ is a normal subgroup of $G$. A morphism $(H,G) \to (H',G')$ is a homomorphism of groups $f : G \to G'$ with $f(H) \subseteq H'$. There is a functor $$Q : \mathbf{Grp}_{\mathrm{sub}} \to \mathbf{Grp},~ (H,G) \mapsto G/H$$ with the evident definition on morphisms: any homomorphism $f : G \to G'$ with $f(H) \subseteq H'$ extends to a homomorphism $[f] : G/H \to G'/H'$, defined by $[f] \circ \pi_H = \pi_{H'} \circ f$.

So the fundamental property shows that $(H,G) \cong (H',G')$ implies $G/H \cong G'/H'$. Here, the relation $(H,G) \cong (H',G')$ means precisely that there is an isomorphism $f : G \to G'$ with $f(H) = H'$. This is the correct replacement of the counterexample mentioned in the beginning.

And this special case is also useful in practice. To give a simple example, consider the isomorphism $2 : \mathbb{Z} \to 2\mathbb{Z}$. It maps $2 \mathbb{Z}$ onto $4 \mathbb{Z}$. As a consequence, $\mathbb{Z}/2\mathbb{Z} \cong 2\mathbb{Z}/4\mathbb{Z}$.

Answer 2

You can replace $A$ with an isomorphic object $A'$ if whatever you're talking about does not depend on any other structure.

For example, the reason that two isomorphic subgroups $H,$ $K$ of a group $G$ do not necessarily produce isomorphic quotients $G/H$, $G/K$ is that taking a quotient $G/H$ not only depends on the group $H$ but also on the map $H\to G.$ The groups $H$ and $K$ are isomorphic abstractly, say to a group $H',$ but the compositions $f:H'\to H\to G$ and $g:H'\to K\to G$ are different, so you should expect the quotients to be different in general. When you study category theory you can check that this means that the morphisms $H\to G$ and $K\to G$ are not isomorphic (in whatever category), even though $H$ and $K$ are, and those morphisms are really the input data to make a quotient group.

In the direct sum example, the data to construct $A\oplus B$ depends only on $A$ and $B$ and nothing else, so replacing them with isomorphic objects doesn't change anything, i.e. the resulting object will be isomorphic to $A\oplus B.$

So this is all to say that one needs to be careful about what data is needed to make some construction. If you want to change the data to some new data, then you need to make sure that the two sets of data are isomorphic to one another, not just part of these sets of data.

Answer 3

Quotienting should be thought of as a process associated to a pair $G$ and a specific subset $H$ (satisfying the condition of being a normal subgroup). Quotient construction depends on the interrelationship between $G$ and $H$; so saying $H$ is a subgroup of $G$ does not capture the relationship fully. So replacing $H$ by another subgroup which is isomorphic to it.

For a rough analogy (not a perfect one), imagine a 5-sided polygon placed in the plane with one of its vertices in the origin, with one side on the x-axis. This presents a particular picture to a viewer. Assume there is another side of the same length, and the remaining three sides all of different lengths. Now 'pull out' the polygon from the plane and fix it back with the other equal side on the x-axis. This will now present a different view.

So quotienting captures how an object is embedded in a bigger object.

In a dihedral group of order 8, there are 5 subgroups of order 2, but only one of them is a normal subgroup. SO quotienting by other, does not even make sense.

Even in the abelian case, group of invertible elements of Z modulo 16 has a subgroup of order 2, which is 'direct factor' yielding a cyclic quotient and another isomorphic subgroup will yield Klein's group.