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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 $\phi(G)$.

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

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

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