Given groups *G* and *H*, recall we have the product *P* = *G* × *H* and projection maps *π*_{1}:*P* → *G* and *π*_{2}:*P* → *H*. There’s nothing mysterious about the projection maps: these just take (*g*, *h*) to *g* and *h* respectively. But what really makes (*P*, *π*_{1}, *π*_{2}) different from all other such triplets is that in a sense, it’s the “boss” of them all.

Let

G,Hbe groups. AproductofGandHis a triplet (P,π_{1},π_{2}), wherePis a group,π_{1}:P→Gandπ_{2}:P→Hare group homomorphisms such that:

- for any triplet (
Q,σ_{1}:Q→Gandσ_{2}:Q→H) there is auniquemapf:Q→Psuch thatπ_{1}f=σ_{1}andπ_{2}f=σ_{2}.We call this the

universal propertyof the product.

Indeed, our (* G × H*,

*π*

_{1},

*π*

_{2}) satisfies this property since

*f*must take

*x*in

*Q*to (

*σ*

_{1}(

*x*),

*σ*

_{2}(

*x*)) in

*G*×

*H*, no choice about it. So our

*G*×

*H*does satisfy the above universal property. And as the Chinese saying goes, you can’t have two tigers on the same mountain:

Any two products (

P,π_{1},π_{2}) and (P’,π’_{1},π’_{2}) are isomorphic, in the sense that there is a unique isomorphismf:P’→Psuch thatπ_{1}f=π’_{1}andπ_{2}f=π’_{2}.

Before we begin the proof, note that existence of *f* is guaranteed by the universal property; we only need to show it’s an isomorphism.

*Proof*: First, apply the universal property to (*P*, *π*_{1}, *π*_{2}) itself, and we see that identity map *h = *1* _{P}* is the only

*P*→

*P*map for which

*π*

_{1}

*h*=

*π*_{1}and

*π*

_{2}

*h*=

*π*_{2}.

Next, take product (*P*, *π*_{1}, *π*_{2}) and apply universal property to the triplet (*Q*, *σ*_{1}, *σ*_{2}) = (*P’*, *π’*_{1}, *π’*_{2}). This induces a unique map *f* : *P’* → *P* such that *π*_{1}*f* = *π’*_{1} and *π*_{2}*f* = *π’*_{2}.

Do the same with product (*P’*, *π’*_{1}, *π’*_{2}) and apply universal property to (*P*, *π*_{1}, *π*_{2}); we get *g* : *P* → *P’* such that *π’*_{1}*g* = *π*_{1} and *π’*_{2}*g* = *π*_{2}.

Thus, *π*_{1}*fg* = *π*_{1 }and *π*_{2}*fg* = *π*_{2}; the first statement tells us *fg* = 1* _{P}*. Likewise, since

*π’*

_{1}

*gf*=

*π’*

_{1}and

*π’*

_{2}

*gf*=

*π*

_{2}, we conclude similarly that

*gf*= 1

*. Thus*

_{P’}*f*and

*g*are isomorphisms and mutual inverses.

–

Ok, time out! The reader may have a foreboding that there’s much more to the above argument than mere cute abstraction. Yes, everyone knows what a group product looks like, and there’s little need to raise it to such an abstract level if that’s all we care about. But the above reasoning is powerful in the sense that *we never took any elements from the groups*. Instead, we’re merely tracing arrows around, composing them, and arguing based on sheer logic. There’re two advantages to this.

- The same concept may be repeated for various algebraic structures, e.g. finite products exist for sets, rings, vector spaces, modules, topological spaces, Hilbert spaces, Banach spaces, homological sequences, sheaves, schemes, etc. It’s nice to do the above work once and stick to it.
- For some algebraic structures like sheaves and schemes, the definition is so involved that directly manipulating the “elements” (if there were any to speak of) is like swimming through butter. The above approach is so much neater.

Trust us: once you start learning schemes, you’ll be thanking your lucky stars for such abstraction.

–

Let’s continue our abstract nonsense: denoting the (abstract) product by *G* × *H* now, prove the following as an exercise.

For any groups *G*, *H*, *K*, *L*,

- ;
- ;
- if
:*g*_{1 }*G*→*K*and:*g*_{2 }*H*→*L*are maps, then we get a unique*g*:*G×H*→*K×L*such that*π*_{K}*g*=*g*_{1}*π*_{G}and*π*_{L}*g*=*g*_{2}*π*_{H}where the projection maps are:

[ Hint for the first two problems: prove that the RHS satisfies the universal property for the LHS, or vice versa ]