Identity elements in a cyclic group Let's say we have  $C_8 = \langle x\rangle$ and $C_6 = \langle y \rangle$. We define $\varphi:C_8 \rightarrow C_6$ by $\varphi(x)=y^3$. I have to find $ker(\varphi)$ and $im(\varphi)$.
So in this case $ker(\varphi) = \{x \in C_8: \varphi(x) = e'\} $ and
$im(\varphi) = \{\varphi(x): x \in C_8\}$ 
Here's what I have so far: 
We know that the $\ker(\varphi)$ is all the elements in $C_8$ that map to the identity of $C_6$,
\begin{align*}
\text{Now an element } x^k \in ker(\varphi) &\iff \varphi(x^k)=1\iff \\
(\varphi(x))^k=1 &\iff n|k 
\end{align*}
(for n being the order of $C_6$). 
Thank you so much in advance!

If $\varphi(x)=y^3$, then $\varphi(x^2)=\varphi(x)^2=y^6=e'$
From this, it's clear that every odd power of $x$ maps to $y^3$, and every even power of $x$ maps to $e'$. (If $n$ is odd, $n=2k+1$, then $\varphi(x^n)=\varphi(x)^{2k+1}=(y^3)^{2k}y^3=(e')^ky^3=y^3$; argue similarly if $n$ is even, but without the extra $y^3$.)
Do you see how this is sufficient for determining the kernel and image of the map?
