My question is, for |G|=30, where G is a cyclic group of order $n$ where $G=<a>$. Consider the mapping from $\phi :G \rightarrow G'$ by $\phi(a^k)=b^k$.

I have showed it is a homomorphism and well defined.

Now Suppose |G|=30 and and |b|=6 for $b\in G'$ and the order of b divides |a|=n. What is the ker($\phi$) and what does the Fundamental Isomorphism Thm say in this case?

Sidebar: Im assuming by Fundamental, they mean first?

My main question is what is the kernal in this case? Assuming they are talking about the First Isomorphism Thm, $\frac{G}{ker(\phi)}$ is isomorphic to $\phi(G)$.


$Ker(\varphi) = \{x: \varphi(x) = e\}$. Then for what $k$ $\varphi(a^k) = \varphi^k(a) = b^k = e$?

  • $\begingroup$ a^n or any multiple of n right? $\endgroup$ – Jack Armstrong Nov 29 '14 at 16:45

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.