Homomorphisms from $\mathbb{Z}_{20}$ to $\mathbb{Z}_8$ I am trying to find all homomorphisms from $\mathbb{Z}_{20}$ to $\mathbb{Z}_8$. I understand how to do it - one completely determines any homomorphism, say $\phi$, by computing multiples of $\phi(1)$ or more generally $\phi(g)$ where $g$ generates $\mathbb{Z}_{20}$. Call $\phi_i$ the mapping taking $1$ to $i$ in $\mathbb{Z}_8$. I don’t know how else to determine whether each $\phi_i$ is a homomorphism besides going through the painstaking process of calculating each $\phi_i(x+y)$ and comparing it to $\phi_i(x)+\phi_i(y)$.
Is there a way to easily rule out certain $\phi_i$s?
Any help would be greatly appreciated.
 A: You are correct about looking at $\phi(1)$: since $1$ generates the domain (i.e. $\mathbb Z_{20}$), you can just look at where $1$ gets sent under the homomorphism. The only other thing you need to check is that the homomorphism preserves the relations. There is one relation in $\mathbb Z_{20}$, namely $20=0$. So $\phi$ better send $20$ to $0$ in $\mathbb Z_8$ (i.e. a multiple of $8$). For example, if you have a map which sends $1$ to $3$, then this would send $20$ to $60$, which would mean it's not a homomorphism (since $60$ is not $0$ in $\mathbb Z_8$).
The real reason behind this is the universal property of quotients. Basically you're looking at group homomorphisms $\mathbb Z\to\mathbb Z_8$ (which are completely determined by where $1$ is sent, and you can send $1$ anywhere) and then asking that the kernel contains the ideal $20\mathbb Z\subseteq\mathbb Z$ so that it factors as a map $\mathbb Z_{20}\to\mathbb Z_8$.
A: We need to find how many $i \in \mathbb{Z}_{8}$ satisfy the relation $20i=0$. There are only $8$ cases to consider:

*

*$i=0$ works ($0$ is divisible by $8$)

*$i=1$ doesn't work ($20$ is not divisible by $8$)

*$i=2$ works ($40$ is divisible by $8$)

*$i=3$ doesn't work ($60$ is not divisible by $8$)

*$i=4$ works ($80$ is divisible by $8$)

*$i=5$ doesn't work ($100$ is not divisible by $8$)

*$i=6$ works ($120$ is divisible by $8$)

*$i=7$ doesn't work ($140$ is not divisible by $8$)

So, there are exactly $4$ group homomorphisms from $\mathbb{Z}_{20}$ to $\mathbb{Z}_{8}$. Only even values of $i$ work, and the odd values can be ruled out.
A: By the First Homomorphism Theorem, the quotient $\Bbb Z_{20}/\operatorname{ker}\phi$ is isomorphic to the image of $\phi$. Since $|\operatorname{im}\phi|$ must be a common divisor of $20$ and $8$, the only possible options for the image of $\phi$ are:

*

*$\operatorname{im}\phi=\{0\}$;

*$\operatorname{im}\phi\cong \Bbb Z_2$;

*$\operatorname{im}\phi\cong \Bbb Z_4$.

Option 1 corresponds to the trivial homomorphism: all the elements of $\Bbb Z_{20}$ are sent to $0$.
Option 2 would correspond to $\operatorname{ker}\phi\cong \Bbb Z_{10}$: $\operatorname{ker}\phi$ is sent to $0$, and $1+\operatorname{ker}\phi$ (which has order $2$ in the quotient) is sent to $4$. You can check that this is indeed a homomorphism.
Option 3 would correspond to $\operatorname{ker}\phi\cong \Bbb Z_{5}$:

*

*$\operatorname{ker}\phi$ is sent to $0$;

*$1+\operatorname{ker}\phi$ (which has order $4$ in the quotient) is sent to either $2$ or $6$;

*$2+\operatorname{ker}\phi$ (which has order $2$ in the quotient) is sent to $4$;

*$3+\operatorname{ker}\phi$ (which has order $4$ in the quotient) is sent to either $6$ or $2$, according to where $1+\operatorname{ker}\phi$ was sent to before.

You can check that both these are indeed homomorphisms.
