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!
 A: 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?
A: One way, but perhaps not the most enlightening way, is just to calculate $\phi(g)$ for every $g \in C_8 = \langle x\rangle$.
So:
$\phi(e) = \phi(x^0) = (y^3)^0 = y^0 = e'\\
\phi(x) = y^3\\
\phi(x^2) = [\phi(x)]^2 = (y^3)^2 = y^6 = e'\\
\phi(x^3) = (y^3)^3 = y^9 = y^6y^3 = y^3\\
\phi(x^4) = (y^3)^4 = y^{12} = (y^6)^2 = (e')^2 = e'\\
\phi(x^5) = (y^3)^5 = y^{15} = y^{12}y^3 = y^3\\
\phi(x^6) = (y^3)^6 = y^{18} = (y^6)^3 = (e')^3 = e'\\
\phi(x^7) = (y^3)^7 = y^{21} = y^{18}y^3 = y^3.$
This makes it clear that $\text{ker }\phi = \{e,x^2,x^4,x^6\}$, and $\text{im }\phi = \{e',y^3\}$. So that gives us the answer, but it doesn't give any clues as to how we might find the answer with different cyclic groups.
If we attempt to use your approach to find the kernel, we have:
$\phi(x^k) = e' \implies y^{3k} = e' \implies 6|3k$.
So this gives us $2|k$, that is, $k$ must be even. But that is only half the story: although we now know with confidence:
$\text{ker }\phi \subseteq \{e,x^2,x^4,x^6\}$, we haven't established that the kernel isn't actually smaller than this set.
However, if we know that a kernel is a subgroup (which you should know), and that any subgroup of a cyclic group is also cyclic, we have three choices for the kernel:
$\{e\} (= \langle e\rangle)\\
\{e,x^4\} (= \langle x^4\rangle)\\
\{e,x^2,x^4,x^6\} = (\langle x^2\rangle).$
You have wisely eliminated the first possibility, by noting $\phi$ is not injective (since $8 > 6$). The answer above shows you $x^2 \in \text{ker }\phi$, which then settles the matter.
From our determination of the kernel, we know exactly half the elements of $C_8$ are mapped to the identity of $C_6$. So the image (which is a subgroup of $C_6$) has at least two elements, and therefore is either:
$\{e',y^3\} = \langle y^3\rangle$, $\{e',y^2,y^4\} = \langle y^2\rangle$, or $C_6$.
Since $y^3 = \phi(x) \in \text{im }\phi$, the second option is out. Since we can write , for any odd  $k$, $k = 2t+1$,
it is clear that for such an odd $k$:
$\phi(x^k) = \phi(x^{2t+1}) = \phi(x^{2t})\phi(x) = \phi((x^2)^t)\phi(x) = [\phi(x^2)]^t\phi(x) = (e')^t\phi(x) = \phi(x) = y^3$.
Note that:
$|\text{im }\phi| = 2 = \dfrac{8}{4} = \dfrac{|C_8|}{|\text{ker }\phi|}$. This is no accident. It is also no accident that $C_8$ splits cleanly into those elements that map to $e'$, and those that map to $y^3$ (and that each such set has exactly the same size-because these are cosets).
