Let $A=\langle a,b: a-3b=0,3a=3b \rangle$ and $B=\mathbb Z^3/S$ with $S=\{m \in \mathbb Z^3: m_1+2m_2+m_3=0, 5|m_3 \}$. Calculate $\operatorname{Hom}_{\mathbb Z}(A,B)$.

My attempt at a solution

I am having some doubts related to the problem so I'll write what I've done so far, I would be greatly appreciated if someone could help me to finish the problem and make any necessary remarks or corrections.

So I am given two finite abelian groups by their relations and generators and they ask me to calculate all the $\mathbb Z$- module morphisms between the two.

We have $A \cong \mathbb Z^2/Im(\phi_A)$, where $\phi_T=\left[ \begin{array}{cc} 1 & 3 \\ -3 & -3\\ \end{array} \right]$

By row operations we get this matrix to its Smith Normal form: $\left[\begin{array}{cc} 2 & 0 \\ 0 & 6\\ \end{array} \right]$

The image is generated by the two columns, so we get $A \cong \mathbb Z^2/\langle (2,0),(0,6) \rangle \cong \mathbb Z/\langle 2 \rangle \oplus \mathbb Z/\langle6 \rangle=\mathbb Z_2 \oplus \mathbb Z_6$

The group $B$ by the relations $m_1+2m_2+m_3=0, 5|m_3$. Notice that this is generated by $m_2(-2,1,0)+k(-5,0,1)$. These are the columns of the matrix $\rho_B=\begin{pmatrix} -2 & -5 \\\\ 1 & 0 \\\\ 0 & 1 \end{pmatrix}$

Its Smith normal form is $\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ 0 & 0 \end{pmatrix}$

So the image of $\rho$ is generated by $(1,0,0),(0,1,0)$. Then $B \cong \mathbb Z^3/\langle (1,0,0),(0,1,0) \rangle \cong \mathbb Z/\langle 1 \rangle \oplus \mathbb Z/\langle 1 \rangle \oplus \mathbb Z/\langle 0 \rangle=\mathbb Z$.

So finally, $\operatorname{Hom}_{\mathbb Z}(A,B)=\operatorname{Hom}_{\mathbb Z}(\mathbb Z_2 \oplus \mathbb Z_6,\mathbb Z)=\operatorname{Hom}_{\mathbb Z}(\mathbb Z_2,\mathbb Z) \oplus \operatorname{Hom}_{\mathbb Z}(\mathbb Z_6,\mathbb Z)$

But any module morphism $f:\mathbb Z_2 \to \mathbb Z$ must satisfy $$0=f(0)=f(2.1)=2f(1)$$ It follows $f(1)=0$, but since $1$ is a generator, then $f=0$. We can arrive to the same conclusion with $\operatorname{Hom}_{\mathbb Z}(\mathbb Z_6,\mathbb Z)$. In conclusion, $\operatorname{Hom}_{\mathbb Z}(A,B)=0$.

  • $\begingroup$ Well, $A$ has two generators and the equations give you relations between these two elements, but I don't understand how you arrive to $A=\{(0,0)\}$ $\endgroup$ – user156441 Dec 10 '14 at 6:07
  • $\begingroup$ The SNF of $A$ is wrong (note that the determinants don't match). It should have entries $1,6$. Then the second column of your matrix for $B$ should be $(-5,0,5)$ not $(-5,0,1)$. $\endgroup$ – Derek Holt Dec 10 '14 at 9:17

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