Kernel of homomorphism between two cyclic groups of diferent order In Malik's abstract algebra one can find the following exercise (and I paraphrase):

Let $f$ be a homomorphism from a cyclic group of order 8 onto a cyclic group of order 4. Determine $\ker f$.

Consider $f: \langle a \rangle_8 \to \langle b \rangle_4$
So: I need to find all the elements $x\in \langle a \rangle_8$ such that $f(x) = e$ ($e \in \langle b \rangle_4$, of course). I can write $x$ as a power of $a$: $x = a^k, k\in \mathbb{Z}$. And then: $$e = f(x) = f(a^k) = \underbrace{f(a)f(a)f(a)\dots f(a)}_{k \mbox{ times, because } f\mbox{ is an homomorphism}} = f(a)^k$$
Now, $f(a) \in \langle b \rangle_4$, $f(a)$ can be then expressed as a power of $b$: $f(a) = b^q$. $$e = f(a)^k = (b^q)^k = b^{qk}$$ Now, I can imply that $4\mid qk$. And this is where my question comes in: because I can assume without loss of generality that $0< q\leq 3$, can I say that $4\mid k$? Because, if I can, I could determine the kernel in the following way: $$\ker f = \{x \in \langle a \rangle_8 \mid x = a^k, 4\mid k\}$$

since the map is surjective the order of the kernel is $\frac84=2$. what subgroup of $<a>$ does this identify?
