Homework 2
Problem 1
Prove there does not exist a functor
Consider a certain sequence of group morphisms
Following the hint, consider the morphism
Now suppose there were some functor
Another way to say this all very succinctly is that group morphisms don't respect membership in centers. Our example is that of the element
Problem 2
Suppose
 Describe the equalizer of
in$A,B:n\to m$ .${\mathbf{\text{Matr}}}_{F}$  Describe the coequalizer of
in$A,B:n\to m$ .${\mathbf{\text{Matr}}}_{F}$
First show that the equalizer should be a matrix

First note that in the category
every commutative diagram of the form${\mathbf{\text{Matr}}}_{F}$ corresponds to an
matrix$n\times l$ such that$D$ . This latter equality is equivalent to the condition that$AD=BD$ , which in turn is equivalent to the condition that$(AB)D=0$ for every column$(AB){\mathbf{d}}_{i}=\mathbf{0}$ of the matrix${\mathbf{d}}_{i}$ . In other words,$D$ exactly when the columns of$AD=BD$ are in the null space of$D$ .$AB$ With that in mind, let
be a basis for the null space of$\{{\mathbf{e}}_{1},\dots ,{\mathbf{e}}_{k}\}$ and let$AB$ be the$E$ matrix with those vectors as its columns. We claim$n\times k$ is an equalizer of the arrows$E:k\to n$ in$A,B:n\to m$ . First note that by construction (and our observation above) we certainly have${\mathrm{Matr}}_{F}$ , i.e., the diagram below commutes:$AE=BE$ Now suppose
is an$D$ matrix such that$n\times l$ , i.e., we have a commutative diagram$AD=BD$ By our earlier observation, this implies that every column
for${\mathbf{d}}_{i}$ is in the null space of$D$ . Since$AB$ is a basis for that null space, there are unique$\{{\mathbf{e}}_{1},\dots ,{\mathbf{e}}_{k}\}$ such that${h}_{j,i}\in F$ . In other words, if we let${\mathbf{d}}_{i}={h}_{1,i}{\mathbf{e}}_{1}+\cdots +{h}_{k,i}{\mathbf{e}}_{k}$ be the corresponding column vector, then${\mathbf{h}}_{i}$ . If we let$E{\mathbf{h}}_{i}={\mathbf{d}}_{i}$ be the matrix with columns$H$ , then we have produced a${\mathbf{h}}_{i}$ matrix$k\times l$ such that$H$ . (Note that this matrix$EH=D$ is also unique determined, since each column$H$ was also uniquely determined.) We've thus produced a unique arrow${\mathbf{h}}_{i}$ through which the arrow$H:l\to k$ factors:$D$ This proves
is an equalizer for the parallel arrows$E:k\to n$ .$A,B:n\to m$ 
This is the dual situation to the previous part. We first simply note that for an
matrix$l\times m$ we have$D$ exactly when$DA=DB$ , or equivalently when$D(AB)=0$ . This in turn is equivalent to every column of$(AB{)}^{\u22ba}{D}^{\u22ba}=0$ (i.e., every row of${D}^{\u22ba}$ ) being in the null space of$D$ . So if we let$(AB{)}^{\u22ba}$ be a$C$ matrix whose rows are a basis for the null space of$k\times n$ , then the same argument as above shows that$(AB{)}^{\u22ba}$ exactly when$DA=DB$ factors uniquely as$D$ . In other words, we have the dual commutative diagram$D=HC$
This proves
Problem 3
Suppose
Recall that an equivalence relation on
The quotient set
Recall that an equivalence relation on
To prove this, first note that for every
Next suppose
Define a set map
This proves
Problem 4
An
 Prove that every finite abelian group is torsion as a
module.$\mathbf{Z}$  Give an example of an infinite abelian group that is torsion as a
module.$\mathbf{Z}$

Suppose
is a finite abelian group, say of order$G$ . By Lagrange's Theorem we then have$n$ for every$ng={0}_{G}$ . But this exactly says that every$g\in G$ is torsion when$g\in G$ is viewed as a$G$ module, and hence$\mathbf{Z}$ is torsion as a$G$ module.$\mathbf{Z}$ 
One nice example is that of the group
of roots of unity in$U$ , although that group is the rare abelian group that uses multiplicative notation. In that group we have (by definition!) that for each$\mathbf{C}$ there is some positive integer$\zeta \in U$ with$n$ . Noting that the${\zeta}^{n}=1$ action on$\mathbf{Z}$ is by$U$ and the "zero element" (i.e., identity) in$m\star \zeta ={\zeta}^{m}$ is 1, we thus have that every$U$ is torsion when$\zeta \in U$ is viewed as a$U$ module.$\mathbf{Z}$ Here's another example. Suppose
is an infinite field of positive characteristic; e.g.,$F$ . Considered as a$F={\mathbf{Z}}_{p}(t)$ module, every element in$\mathbf{Z}$ is killed by the characteristic of the field, and hence is torsion.$F$ A third example: the direct product of an infinite number of copies of a fixed finite abelian group, say
.$G={\displaystyle \prod _{i=1}^{\mathrm{\infty}}{\mathbf{Z}}_{p}}$