What is the number of distinct homomorphism from $\Bbb Z/5 \Bbb Z$ to $\Bbb Z/7 \Bbb Z$ 
What is the number of distinct homomorphism from $\Bbb Z/5 \Bbb Z$  to  $\Bbb Z/7 \Bbb Z$ and how to find it?  

I came across the above problem and do not know how to get it? Can someone point me in the right direction? Thanks in advance for your time.
 A: Hint 1 (looking at subgroups and their orders): Any homomorphism $\Bbb{Z}_5\xrightarrow{\phi}\Bbb{Z}_7$ yields a subgroup of $\Bbb{Z}_7$ given by $\phi\left(\Bbb{Z}_5\right)$. What are the subgroups of $\Bbb{Z}_7$? Given the subgroups and their orders, what possibilites are there for $\phi\left(\Bbb{Z}_5\right)$? (You might want to use Lagrange's theorem.)
Hint 2 (following the hint in the comment): Both groups are cyclic and generated by $1$. Therefore, a homomorphism $\phi : \Bbb{Z}_5\to\Bbb{Z}_7$ is determined by where $1$ is sent (since any element $k\in\Bbb{Z}_5$ is equal to $k\cdot 1 = \underbrace{1 + \ldots + 1}_{k\textrm{ times}}$). So $\phi(1)$ generates a subgroup of $\Bbb{Z}_7$. Think about the possibilities for the size of the subgroup given by $\left<\phi(1)\right>\subseteq\Bbb{Z}_7$ based on the orders of $\Bbb{Z}_5$ and $\Bbb{Z}_7$ (and hence the possibilities for $\phi(1)$).
A: The first homomorphism gives that if $\varphi:\mathbb{Z}_5 \rightarrow \mathbb{Z}_7$ is a homomorphism, then $\mathbb{Z}_5/\operatorname{ker}\varphi\cong \varphi[\mathbb{Z}_5]$.  Since the only subgroups of $\mathbb{Z}_5$ are the trivial subgroup and $\mathbb{Z}_5$ itself, the image of $\varphi$ must either be trivial or isomorphic to $\mathbb{Z}_5$.  No subgroup of $\mathbb{Z}_7$ can be isomorphic to $\mathbb{Z}_5$ by Lagrange's theorem, so $\varphi$ must be the trivial homomorphism.
A: Hint: if $\,G,H\,$ are groups and $\,f:G\to H\,$ a group homomorphism, then
$$(1)\;\;\;\forall\,g\in G\;,\;\;ord(f(g))\mid ord(g)$$
$$(2)\;\;\;\text{Both groups given are cyclic of (co)prime order}\ldots$$
A: I feel like the other answers have trouble getting to the point.   The only subgroups of $\mathbb{Z}_7$ are the trivial group and the entire group since $7$ is prime. Since $\phi(\mathbb{Z}_5) \le \mathbb{Z}_7$ is a subgroup, it must have either $1$ or $7$ elements.  It can't possibly have $7$ since $\mathbb{Z}_5$ is a group of order $5$.  So $\phi(\mathbb{Z}_5) = \{0\} \implies$ $\phi$ is trivial.
A: Let group hom $\rm\:f\,:\,\Bbb Z_5\to \Bbb Z_7\:$ with kernel $\rm\,K,\,$ and image $\rm\,I\,$ of size $\rm\:\#I = n$.   
Note $\rm\ n\mid 7\ $ by Lagrange: $\rm\ I\,$ is a subgroup of $\rm\,\Bbb Z_7\:\Rightarrow\: \#I \mid \#\Bbb Z_7,\ $ i.e. $\rm\ n\mid 7$.   
Note $\rm\ n\mid 5\ $ by $\rm\ I \cong \Bbb Z_5/K\:\Rightarrow\: \#I\, =\, \#\Bbb Z_5/\#K,\ $ i.e. $\rm\ n = 5/\#K$
Thus $\rm\:n\mid 5,7\:\Rightarrow\:n\mid gcd(5,7)=1,\ $ i.e. $\rm\ \#I = n = 1,\:$ so the image $\rm\,I\,$ is trivial. 
