1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $K_s$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.