Show the following:
For any abelian group \(A\) and any positive integer \(m\), \[ \mathrm{Hom}(\mathbb{Z}_m, A) \cong A[m] \coloneqq \{ a\in A \mid ma = 0\} .\]
\(\mathrm{Hom}(\mathbb{Z}_m, \mathbb{Z}_n) \cong \mathbb{Z}_{\mathrm{gcd}(m,n)}\).
As a \(\mathbb{Z}-\)module, \(\mathbb{Z}_m^* = 0\).
For each \(k\geq 1\), \(\mathbb{Z}_m\) is a \(\mathbb{Z}_{mk}-\)module, and as a \(\mathbb{Z}_{mk}\) module, \(\mathbb{Z}_m^* \cong \mathbb{Z}_m\).
Solution
1. (1) The map \(\Phi\colon \operatorname{Hom}(\ZZ_m, A) \to A[m]\), \(\Phi(f) = f(\bar 1)\), is an isomorphism.
1.1. \(\Phi\) is a well-defined homomorphism.
For \(f\in\operatorname{Hom}(\ZZ_m,A)\), \(m f(\bar 1) = f(m \bar 1) = f(\bar 0) = 0\), so \(f(\bar 1) \in A[m]\). Also \(\Phi(f + g) = f(\bar 1) + g(\bar 1) = \Phi(f) + \Phi(g)\).
1.2. \(\Phi\) is injective.
If \(f(\bar 1) = 0\), then \(f(\bar k) = k f(\bar 1) = 0\) for every \(k\), so \(f = 0\).
1.3. \(\Phi\) is surjective.
Given \(a \in A[m]\), put \(f(\bar k) = ka\). If \(\bar k = \bar k'\), then \(k - k' = qm\) for some \(q\in\ZZ\), so \(ka - k'a = q(ma) = 0\); hence \(f\) is a well-defined homomorphism with \(\Phi(f) = a\).
1.4. Q.E.D.
Steps <2>1–<2>3.
2. (2) \(\operatorname{Hom}(\ZZ_m, \ZZ_n) \cong \ZZ_{\gcd(m,n)}\).
By step 1 with \(A = \ZZ_n\), \(\operatorname{Hom}(\ZZ_m, \ZZ_n) \cong \ZZ_n[m]=\{\bar a \in \ZZ_n : ma \equiv 0 \pmod n\}\). Put \(d=\gcd(m,n)\). Then \(n\mid ma\) if and only if \(\frac nd\mid \frac md a\), if and only if \(\frac nd\mid a\), since \(\gcd(\frac md,\frac nd)=1\). So \(\ZZ_n[m]\) is the subgroup of \(\ZZ_n\) generated by \(\overline{n/d}\), which is cyclic of order \(d\).
3. (3) \(\ZZ_m^* = \operatorname{Hom}_\ZZ(\ZZ_m, \ZZ) = 0\).
By step 1, \(\operatorname{Hom}(\ZZ_m, \ZZ) \cong \ZZ[m] = \{a \in \ZZ : ma = 0\}\), which is \(0\) because \(\ZZ\) is torsion-free.
4. (4) \(\ZZ_m\) is a \(\ZZ_{mk}\)-module via \(\bar r \cdot \bar a = \overline{ra}\), and \(\operatorname{Hom}_{\ZZ_{mk}}(\ZZ_m, \ZZ_{mk}) \cong \ZZ_m\).
4.1. The action \(\bar r \cdot \bar a = \overline{ra}\), for \(\bar r \in \ZZ_{mk}\) and \(\bar a \in \ZZ_m\), is well defined.
If \(r\equiv r'\pmod{mk}\), then \(r-r'\) is a multiple of \(m\), so \(ra\equiv r'a\pmod m\); the module axioms are those of \(\ZZ_m\) as a \(\ZZ\)-module.
4.2. A \(\ZZ_{mk}\)-module homomorphism \(f\colon \ZZ_m \to \ZZ_{mk}\) is determined by \(b=f(\bar 1)\), and the possible values are exactly the \(b \in \ZZ_{mk}\) with \(mb = 0\).
Since \(\bar 1\) generates \(\ZZ_m\), \(f\) is determined by \(f(\bar 1)\), and \(m\bar 1 = 0\) forces \(m f(\bar 1) = 0\). Conversely, for \(b\) with \(mb=0\), the map \(\bar a\mapsto ab\) is well defined and \(\ZZ_{mk}\)-linear, as in step 3 of step 1.
4.3. The elements \(b \in \ZZ_{mk}\) with \(mb = 0\) form the subgroup \(\{0, \bar k, 2\bar k, \ldots, (m-1)\bar k\}\), which is isomorphic to \(\ZZ_m\) as a \(\ZZ_{mk}\)-module.
\(mb \equiv 0 \pmod{mk}\) if and only if \(k\mid b\). The multiples of \(\bar k\) form the cyclic submodule generated by \(\bar k\), which has order \(m\), and \(\bar a\mapsto a\bar k\) is a \(\ZZ_{mk}\)-linear isomorphism \(\ZZ_m\to\langle\bar k\rangle\).
4.4. Q.E.D.
By steps 2 and 3, \(f\mapsto f(\bar 1)\) is a \(\ZZ_{mk}\)-linear isomorphism \(\ZZ_m^*=\operatorname{Hom}_{\ZZ_{mk}}(\ZZ_m, \ZZ_{mk})\to\langle\bar k\rangle\cong\ZZ_m\).