# Notes:Products and sums of groups

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## Recall

Given a pair of groups, [ilmath](A,*_A)[/ilmath] and [ilmath](B,*_B)[/ilmath] in the category of groups we define the following:

### Product

• A wedge: $\xymatrix{ A \\ S \ar[u]_{p_A} \ar[d]^{p_B} \\ B }$ such that for each wedge $\xymatrix{ & A \\ X \ar[ur]^{f_A} \ar[dr]_{f_B} & \\ & B}$ there exists a unique arrow [ilmath]\xymatrix{&\\X \ar[r]^m & S\\&} [/ilmath] such that $\xymatrix{ & A \\ X \ar[r]^m \ar[ur]^{f_A} \ar[dr]_{f_B} & S \ar[u]_{p_A} \ar[d]^{p_B} \\ & B}$ commutes.

It is obvious that

### Coproduct (sum)

• A wedge: $\xymatrix{ A \\ S \ar@{<-}[u]^{i_A} \ar@{<-}[d]_{i_B} \\ B }$ such that for each wedge $\xymatrix{ A & \\ & X \ar@{<-}[ul]_{f_A} \ar@{<-}[dl]^{f_B} & \\ B&}$ there exists a unique arrow [ilmath]\xymatrix{&\\S \ar[r]^m & X\\&} [/ilmath] such that $\xymatrix{A \ar[dr]^{f_A} \ar[d]_{i_A}& \\ S \ar[r]^m & X \\ B \ar[u]^{i_B} \ar[ur]_{f_B} &}$ commutes.

## "Proof" that [ilmath](A\times B,*_{A\times B})[/ilmath] is the product and coproduct of [ilmath]A[/ilmath] and [ilmath]B[/ilmath] (FALSE)

Here we define the operation: [ilmath]*_{A\times B}:(A\times B)\times(A\times B)\rightarrow A\times B[/ilmath] as follows:

• [ilmath]*_{A\times B}:((a,b),(a',b'))\mapsto(aa',bb')[/ilmath]

(This means that I am claiming [ilmath]S=A\times B[/ilmath] with this operation)

### Product (TRUE)

Let [ilmath](X,\cdot)[/ilmath] be a group and let [ilmath]f_A:X\rightarrow A[/ilmath] and [ilmath]f_B:X\rightarrow B[/ilmath] be group homomorphisms be given.

• We must find a unique [ilmath]m:X\rightarrow A\times B[/ilmath] (also a group homomorphism) such that:
1. [ilmath]p_A\circ m=f_A[/ilmath] and
2. [ilmath]p_B\circ m=f_B[/ilmath]

I claim that [ilmath]A\times B[/ilmath] is a group with this operation, and, [ilmath]p_A[/ilmath] and [ilmath]p_B[/ilmath] are defined as follows (and are group homomorphisms):

• [ilmath]p_A:A\times B\rightarrow A[/ilmath] by [ilmath]p_A:(a,b)\mapsto a[/ilmath] and
• [ilmath]p_B:A\times B\rightarrow B[/ilmath] by [ilmath]p_B:(a,b)\mapsto b[/ilmath]

In this event, [ilmath]A\times B[/ilmath] is the categorical product of [ilmath]A[/ilmath] and [ilmath]B[/ilmath].

• Clearly we can define [ilmath]m:X\rightarrow A\times B[/ilmath] by [ilmath]m:x\mapsto(f_A(x),f_B(x))[/ilmath]

Then [ilmath]p_A(m(x))=p_A((f_A(x),f_B(x))=f_A(x)[/ilmath] as required.

### Coproduct (FALSE)

Let [ilmath](X,\cdot)[/ilmath] be a group and let [ilmath]f_A:A\rightarrow X[/ilmath] and [ilmath]f_B:B\rightarrow X[/ilmath] be group homomorphisms be given.

• We must find a unique [ilmath]m:A\times B\rightarrow X[/ilmath] (also a group homomorphism) such that:
1. [ilmath]m\circ i_A=f_A[/ilmath] and
2. [ilmath]m\circ i_B=f_B[/ilmath]

I claim that [ilmath]A\times B[/ilmath] is a group with this operation, and, [ilmath]i_A[/ilmath] and [ilmath]i_B[/ilmath] are defined as follows (and are group homomorphisms):

• [ilmath]i_A:A\rightarrow A\times B[/ilmath] by [ilmath]i_A:a\mapsto(a,0)[/ilmath] and
• [ilmath]i_B:B\rightarrow A\times B[/ilmath] by [ilmath]i_B:b\mapsto(0,b)[/ilmath]

In this event [ilmath]A\times B[/ilmath] is the categorical coproduct of [ilmath]A[/ilmath] and [ilmath]B[/ilmath]

• After a little thought I've come up with:
• [ilmath]m:A\times B\rightarrow X[/ilmath] given by [ilmath]m:(a,b)\mapsto f_A(a)\cdot f_B(b)[/ilmath] HOWEVER is is clear that
• [ilmath]m:(a,b)\mapsto f_B(b)\cdot f_A(a)[/ilmath] would also work just as well; and is distinct from the [ilmath]m[/ilmath] above unless [ilmath]X[/ilmath] is an Abelian group.
• As [ilmath]m(i_A(a))=m((a,0))=f_A(a)\cdot f_B(0)=f_A(a)\cdot e_B=f_A(a)[/ilmath] as required, or:
• [ilmath]m(i_A(a))=m((a,0))= f_B(0)\cdot f_A(a)=e_B\cdot f_A(a)=f_A(a)[/ilmath] depending on how you define [ilmath]m[/ilmath], similarly for [ilmath]f_B[/ilmath]
• Thus we see that (unless the choice of [ilmath]X[/ilmath] is always an Abelian group) that there is no unique mediating arrow, and thus [ilmath]A\times B[/ilmath] cannot be the coproduct of [ilmath]A[/ilmath] and [ilmath]B[/ilmath].