Prove that $\phi : (\mathbb Z/25, +)\to (\mathbb Z/5, +)$ such that $\phi([k]) = [k].$ is a well-defined homomorphism. 
(1) Define $\phi : (\mathbb Z/25, +)\to (\mathbb Z/5, +)$ by $\phi([k]) = [k].$ Prove that $\phi$ is actually well-defined and is a homomorphism.

To show that $f$ is well defined, we need to show that whenever $[a]_{25} = [b]_{25}$
in $\mathbb{Z_{25}}$, then $f([a]_{25}) = f([b]_{25})$ in $\mathbb{Z_5}$. If $[a]_{25} = [b]_{25}$, then $a − b = 25t$ for
some $t ∈ \mathbb{Z}$. Thus $a − b = 5(5t)$ and hence $f([a]_{25}) = [a]_5 = [b]_5 = f([b]_{25})$
as required.
To show that $f$ is a homomorphism, note that for any $[a]_{25}, [b]_{25} ∈ \mathbb{Z_{25}}$,
$f([a]_{25} +[b]_{25}) = f([a+b]_{25}) = [a+b]_{5} = [a]_{5} +[b]_{5} = f([a]_{25})+f([b]_{25})$ and
$f([a]_{25}[b]_{25}) = f([ab]_{25}) = [ab]_5 = [a]_5[b]_5 = f([a]_{25})f([b]_{25})$ as required.
To show that $f$ is surjective note that $f([0]_{25}) = [0]_5, f([1]_{25}) = [1]_5, f([2]_{25}) =
[2]_5$, and $f([3]_{25}) = [3]_5$. Because we can hit everything in $\mathbb{Z_5}$, $f$ is a
surjection.

2) Find $\ker(\phi).$

An element $[a]_{25} \in \mathbb{Z_{25}}$ is in the kernel of $f$ if and only if $[a]_5 = [0]_5$, that is, if $5 | a$. The integers between $0$ and $25$ which are divisible by $5$ are $\{0, 5, 10, 15, 20\}$. So the kernel of f is the ideal generated by $\{0, 5, 10, 15, 20\}$.
I believe this is the solution but I'm not sure exactly how to read this.
 A: This is mostly good, but a few things to nit-pick.
The notation indicates only that this is a group homomorphism. If the problem wanted ring homomorphisms, it would not specify $(\mathbb Z_{25},+)$ as the structure. It would say either $(\mathbb Z_{25},+,\cdot)$ or, if it was implicit in the context that it was about ring homomorphisms, it might leave off the notation entirely, and simply write $\mathbb Z_{25}\to\mathbb Z_{5}.$
So you don't need to show $f(ab)=f(a)f(b),$ since that's only needed for a ring homomorphisms. In this case, the homomorphism is also a ring homomorphism, so it doesn't hurt you. But there are group homomorphism $(\mathbb Z_{25},+)\to(\mathbb Z_{5},+)$ which are not ring homomorphisms, like $[k]_{25}\mapsto [2k]_{5}.$
Because of this, the kernel of the homomorphism of groups is not, in general, called an ideal. Here, it is an ideal in some extension of this group to a ring, but that is incidental. Here, "subgroup" is better.
Also, since the kernel is a subset of $\mathbb Z_{25},$ it can't really be say it can be "generated by" $\{0,5,10,15,20\},$ because these are not elements of $\mathbb Z_{25}.$ That's nit-picky, but since we are being careful about equivalence classes... 
While technically true that the kernel is "generated by" $K=\{[0]_{25},[5]_{25},[10]_{25},[15]_{25},[20]_{25}\},$ you can also say more strongly that the kernel is equal to this set $K,$ and is generated by $[5]_{25}$.  
