Order of kernel of a homomorphism. Let $C_n$ denotes the cyclic group of order $n$ and let  $\phi:C_{52}\rightarrow C_{52}$ be the homomorphism $\phi(x)=x^7$. What is the order of kernel of  $\phi$?  
I know that $ker\phi=\left\{x/ x^7=e\;;\; e \;\;\text{is the identity in}\;\;  C_{52}\ \right\}$.  
Is there any general formula to find for kernel of $C_n$?
Thank you.
 A: In fact, if we have $\phi(x)=x^m$ acting on the group $C_n$, we know that $\lvert \ker(\phi) \rvert\mid n$ since $\ker(\phi)\leq C_n$. We also know that $\ker(\phi)\leq C_n$ is cyclic. So there exists some element $k\in\ker(\phi)$ such that $\langle k\rangle=\ker(\phi)$. Since $k^m=e$ we know that $o(k) \mid m$ and hence, $o(k)= \lvert \ker(\phi) \rvert \mid m$.
Now suppose that $l \mid m,n$ for some $l\in \mathbb{Z}$. Since $l \mid m$ we can say that there exists some element $x\in \ker(\phi)$ where $o(x)=l$ (note that since $l \mid n$, $l\leq n$). Because $\ker(\phi)\leq C_n$ we know that it is closed and that $\langle x\rangle\leq \ker(\phi)$. So $l=o(x)=\lvert\langle x\rangle\rvert\mid \lvert\ker(\phi)\rvert$. So if $l \mid n,m$ then $l\mid \lvert \ker(\phi) \rvert$. So if some integer divides both $n$ and $m$ it also divides $\lvert \ker(\phi) \rvert$. Hence, $\lvert \ker(\phi) \rvert=(m,n)$.
A: The kernel of $x^r : C_n \to C_n$ is going to be all those elements $x$ of $C_n$ for which $x^r = 1$.
If $r \mid n$ let $k = n/r$ and we have the subgroup $C_r \le C_n$ by the inclusion $i \mapsto ki : C_r \to C_n$ (I am identifying the cyclic groups with modular arithmetic) and this is exactly the kernel we were interested in.
If $r \not \mid n$ then there are no elements (other than the identity) for which $x^r = 1$, since otherwise one of them would generate a $C_r \le C_n$ which is a contradiction.

Here is an example.


*

*$C_{14}$: 0,1,2,3,4,5,6,7,8,9,10,11,12,13

*The $C_{7}$ subgroup: 0,2,4,6,8,10,12


So we see there are 7 elements in the kernel of $x^7$, they are the squares.
