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Suppose $G$ is a finite group and $\phi : G \rightarrow \mathbb{Z}_{10}$ is a surjective homomorphism. Then, $G$ normal subgroups of indexes $2$ and $5$.

Approach: Since $\phi$ is surjective then, by first isomoprhism thm, $G/\ker\phi \cong \mathbb{Z}_{10}$. Therefore, $|G| = 10|\ker\phi|$. Also, we know $10 \mid |G|$. Say $N \lhd G$. We need to show $[G:N] = 2$. Now, here I'm stuck, because I don't know how to show that $N$ has index $2$ in G? Can someone help me get to the next step?


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You may want to use the fact that the pre-image of a subgroup is a subgroup, and that $\mathbb Z / (10)$ is abelian... – Niccolò Oct 23 '12 at 23:18
up vote 3 down vote accepted

You are not as far off as you think; you have made the all the right observations thus far.

Write $K=\text{Ker}\,\phi$, as you noted $G/K$ is isomorphic to the cyclic group $\Bbb{Z}_{10}$, so there is a unique subgroup of $G/K$ of index 2. The correspondence theorem then guarantees a subgroup of $G$ of index 2.

I think you can do the rest. Look back at what properties are preserved by the bijection defined by the correspondence theorem if you still feel stuck.

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