# Proving relations between kernels and images of a Group G

Let $G$ be an abelian group and n be an integer. Deﬁne the map $\phi_n\colon G \to G$, $\phi_n(g) = g^n$ since $G$ is abelian $(hg)^n = h^ng^n$ that is $\phi_n$ is a homomorphism. We then have the subgroups $K_n = \operatorname{ker}(\phi_n)$ and $I_n = \operatorname{im}(\phi_n)$. So in other words, we have a map for each integer $n$ that sets up a map from the group to itself. This map 'multiplies' the element by itself $n$ times. The kernel of the map is the maps is the elements in the domain that gives you the identity.

Let G be abelian, |G| = $sp^i$ with $p$ and $s$ relatively prime. Why is it true that: |$K_{p^i}$| = pi and |$K_s$| = $s$? I don't get why it is true. Why is it true on a intuitive level?

Note that $|G|=|K_n|\cdot |I_n|$. An element of order $tp^j$ with $j>0$, when multiplied with $p$, gets order $tp^{j-1}$, whereas multiplication with $p$ cannot alter the order of an element of coprime to $p$ order. Therefore, all elements in the kernel of $\phi_{p^i}$ have order some power of $p$. Thus $|K_{p^i}|$ itself is a power of $p$. On the othre hand, $|I_{p^i}|$ must be coprime to $p$. The only possible factorization is $|K_{p^i}|=p^i$, $|I_{p^i}|=s$.
The case with $n=s$ instead of $n=p^i$ is similar: $I_s$ cannot have element of order $q$ if $q$ s a prime dividing $s$, whereas all elements of order dividing $s$ are certainly in $K_s$.