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The following example is taken from the book "Contemporary Abstract Algebra" by Joseph A. Gallian, seventh edition, page#210.

If $G$ is a group of order $60$ and $G$ has a homomorphic image of order $12$ that is cyclic, then by the properties of subgroups under homomorphism we know that $G$ has normal subgroups of orders $5, 10, 15, 20, 30,$ and $60$.

From the properties given in the book I can deduce the following information

1) $\phi(G)$ is a cyclic subgroup of order $12$ of Co-domain $\bar{G}$

2) $|\phi(G)|$ divides $|G|$ and $|\bar{G}|$

I don't understand how the information about the normal subgroups of $G$ is deduced from a finite cyclic homomorphic image. Am I missing something?

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The hypothesis is that there is a surjective homomorphism $\phi : G \to H$ where $H$ is cyclic of order $12$. If $K$ is the kernel of $\phi$, then $G/K \cong H$.

By the correspondence theorem, there is a bijection between subgroups of $H$ one one hand, and subgroups of $G$ which contain $K$ on the other hand. This bijection preserves indices and normality.

As $H$ is cyclic of order $12$, it has normal (indeed, characteristic) subgroups of all orders dividing $12$, so $1,2,3,4,6,12$. The indices of these subgroups are $12,6,4,3,2,1$ respectively. The corresponding subgroups of $G$ are normal in $G$ and have the same indices, so their orders are $5,10,15,20,30,60$ respectively.

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