Kernel and Image of a map Note: The OP has restored the question to its original form. An edit by someone else inadvertently changed the codomain from $\mathbb Z_{2p}$ (which makes sense) to $\mathbb Z_{2^p}$ (which does not).

Let $p$ be an odd prime. Let $f: \mathbb{Z}_{p^2} \to \mathbb{Z}_{2p}$ be the unique group homomorphism defined by $f(2) = 2$. Determine $ker(f)$ and $image(f)$.
Help is very welcome. I am at a loss. Thank you.
 A: Note: Looking at the edit history, I see that the OP is correct: in the original question, the codomain was $\mathbb Z_{2p}$, but this was changed in an edit (not by the OP) to $\mathbb Z_{2^p}$, which made the problem incorrect as there is no such homomorphism with codomain $\mathbb Z_{2^p}$, as shown in the comment by Jake.
However, the problem as originally posed does have a solution.
First, recall that for any integer $n$ and any positive integer $N$, the notation $[n]_{N}$ refers to the residual class of $n$ modulo $N$.
Note that $[2]_{p^2}$ is a generator for $\mathbb Z_{p^2}$. To see this, suppose that $[2n]_{p^2} = [0]_{p^2}$. This implies that $2n = kp^2$ for some integer $k$. As $p^2$ is odd and $2n$ is even, this means that $k$ must be even, say $k=2m$ for some integer $m$. It follows that $n = mp^2$, which means that $p^2$ divides $n$. Thus the order of $[2]_{p^2}$ must be $p^2$, so $[2]_{p^2}$ is a generator as claimed.
Now define $f : \mathbb Z_{p^2} \to \mathbb Z_{2p}$ by $f([n]_{p^2}) = [n]_{2p}$. Let us verify that $f$ is well-defined. Since $[2]_{p^2}$ generates $\mathbb Z_{p^2}$, two arbitrary elements of $\mathbb Z_{p^2}$ are of the form $[2m]_{p^2}$ and $[2n]_{p^2}$. Suppose these elements are equal. Then $2m - 2n = kp^2$ for some integer $k$. Since the left hand side is even, as above this forces $k=2j$ for some integer $j$. Then $2m - 2n = 2jp^2 = jp(2p)$, so $2m$ and $2m$ differ by an integer multiple of $2p$. Thus $[2m]_{2p} = [2n]_{2p}$, so $f$ is indeed well-defined.
Observe that $f$ satisfies $f([2]_{p^2}) = [2]_{2p}$ as in the problem statement. Let us verify that it is a homomorphism. This is easy:
$$f([m]_{p^2} + [n]_{p^2}) = f([m+n]_{p^2}) = [m+n]_{2p} = [m]_{2p} + [n]_{2p}$$
Now let us find the kernel $K$ of $f$. An arbitrary element $[2n]_{p^2}$ is in $K$ if and only if $f([2n]_{p^2}) = [2n]_{2p} = 0$, which happens if and only if $2n = 2pk$ for some integer $k$, if and only if $n$ is a multiple of $p$. This means that $K = \langle [p]_{p^2} \rangle$, which has order $p$.
As $\mathbb Z_{p^2} / K$ is isomorphic to the image $I$ of $f$, and $|\mathbb Z_{p^2} / K| = |\mathbb Z_{p^2}|/|K| = p^2 / p = p$, this means that $|I| = p$. As $\mathbb Z_{2p}$ is cyclic, it contains a unique subgroup of order $p$, namely $\langle [2]_{2p} \rangle$. It follows that $I = \langle [2]_{2p}\rangle$.
A: The image of $f$ is the subgroup generated by $2$, which clearly has order $p$ in $\mathbb{Z}/2p\mathbb{Z}$. Thus the kernel of $f$ has index $p$ in $\mathbb{Z}/p^2\mathbb{Z}$ and therefore it is $p\mathbb{Z}/p^2\mathbb{Z}$.
Note: we use the fact that a cyclic group of order $n$ has exactly one subgroup of order $m$, where $m\mid n$. The homomorphism is well defined, because $2$ is coprime with $p^2$, so it is a generator of $\mathbb{Z}/p^2\mathbb{Z}$; moreover $p^2\cdot2\equiv0\pmod{2p}$, so the homomorphism is well defined.
