List the number of distinct group homomorphisms from $ \Bbb{Z} / 120 \Bbb{Z} $ to itself. I have trouble finding how many distinct group homomorphisms there are from $ \Bbb{Z} / 120 \Bbb{Z} $ to itself. Does this have something to do with abelian groups of order $ 120 $? Furthermore, if we view $ \Bbb{Z} / 120 \Bbb{Z} $ as a ring, then is its unit group $ (\Bbb{Z} / 120 \Bbb{Z})^{\times} $ a cyclic group? What is the connection between these questions?
 A: Consider a more general problem: finding the endomorphisms of $\def\Z{\mathbb{Z}}\Z/n\Z$, for $n>0$.
Denote by $\pi\colon\Z\to\Z/n\Z$ the canonical projection. If $f\colon\Z/n\Z\to\Z/n\Z$ is a group homomorphism, then $f\circ\pi\colon\Z\to\Z/n\Z$ has a kernel containing $n\Z$; indeed,
$$
f\circ\pi(n)=f(0+n\Z)=0+n\Z
$$
Conversely, if $g\colon\Z\to\Z/n\Z$ is a homomorphism such that $\ker g\supseteq n\Z$, this defines a unique homomorphism $f\colon\Z/n\Z\to\Z/n\Z$ such that $f\circ\pi=g$.
Since $\Z$ is a free group, a homomorphism $g\colon\Z\to\Z/n\Z$ is completely determined by the (arbitrary) image of $1$: $g(z)=zg(1)$.
If $x+n\Z\in\Z/n\Z$, then we can define $g(1)=x+n\Z$ and so $g(z)=zx+n\Z$. Since $g(n)=nx+n\Z=0+n\Z$, we get that any element of $\Z/n\Z$ can be chosen to define $g\colon\Z\to\Z/n\Z$ so that $\ker g\supseteq n\Z$.
Thus the number of endomorphisms of $\Z/n\Z$ is $n$.
A: About the group of units, note that if $x$ is relatively prime to $120$, then $x^4\equiv 1\pmod{8}$, and $x^4\equiv 1\pmod{5}$. Also, $x^2\equiv 1\pmod{3}$, so $x^4\equiv 1\pmod{3}$. It follows that $x^4\equiv 1\pmod{120}$. So every element of the group of units has order $4$ or less, while the group of units has order $32$.
A: $\DeclareMathOperator\Hom{Hom}\Hom_{\mathbf Z}(\mathbf Z/n\mathbf Z,G)\simeq 0:_G 120\mathbf Z$ for any abelian group $G$. Hence
$$\Hom_{\mathbf Z}(\mathbf Z/120\mathbf Z,\mathbf Z/120\mathbf Z)\simeq 0:_{\mathbf Z/120\mathbf Z} 120\mathbf Z\simeq \mathbf Z/120\mathbf Z.$$
The group of units $\;\bigl(\mathbf Z/n\mathbf Z\bigr)^\times$ is cyclic only in the following cases: $n=2^k\ (k\le 2)$ or $n=p^k\ (p$ odd) or $n=2p^k$. So $\;\bigl(\mathbf Z/120\mathbf Z\bigr)^\times$ is not cyclic.
Finally the connection is that $\;\operatorname{Aut}(\mathbf Z/120\mathbf Z)\simeq \bigl(\mathbf Z/120\mathbf Z\bigr)^\times $.
A: $\newcommand{\ZZ}{\mathbb{Z}}$
The key point is that $\ZZ/120\ZZ$ is cyclic. That means that it is generated by one element, namely $1\mod 120$. Certainly defining any homomorphism $\phi$ from $\ZZ/120\ZZ$ to any group $G$ requires knowing where what $\phi(1)$ is in $G$. Since $\ZZ/120\ZZ$ is cyclic, this is actually enough to determine where $\phi$ sends every other element of $\ZZ/120\ZZ$. In your case, setting $G = \ZZ/120\ZZ$, every homomorphism $\phi : \ZZ/120\ZZ\rightarrow\ZZ/120\ZZ$ is uniquely determined by $\phi(1)$.
This is because, if $\phi(1) = n$, then $\phi(2) = \phi(1+1) = \phi(1) + \phi(1) = n + n$. Similarly, $\phi(3) = \phi(1+2) = \phi(1) + \phi(2) = n + n + n$, $\phi(-1) = -\phi(1) = -n$,...etc.
Thus, there are precisely 120 distinct homomorphisms from $\ZZ/120\ZZ$ to itself, corresponding to the 120 possible choices for $\phi(1)$.
