Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thanks!

share|improve this question
    
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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