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If G is cyclic , show |G/H| < ∞ for any subgroup H except the identity.

I already know that any subgroup of a cyclic group is also cyclic but i have no idea how to prove a the quotient of G is finite especially when Lagrange theorem also only apply to finite set.

Thanks everyone for your helps, truly appreciate it!

Minh Vu
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2 Answers2

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I will outline a solution. I think you should be able to fill in the details.

Let $G$ be cyclic and generated by $g$. Then, as we know any subgroup is also cyclic, if $H$ is a subgroup, it is generated by $g^n$ for some $n\in \mathbb N$.

I claim that the index of $H$ in $G$ is $n$, with cosets $H, gH, \cdots, g^{n-1}H$. It suffices to show that any element lies in one of those cosets. Fix an element $g^a$ and find the remainder when $a$ is divided by $n$. Can you see how this tells you which coset it belongs to?

Potato
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  • I was able to tell which coset g^a belongs to but how does it help me with the final answer. Also What if H is infinite? thanks for the help – Minh Vu May 20 '13 at 04:58
  • Knowing that every element belongs to some coset in a finite collection of cosets tells you that the index is finite. Also, my answer handles the case where $H$ is infinite. – Potato May 20 '13 at 05:00
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HINT: Let $g$ be a generator of $G$, and pick $h\in H$ with $h\ne 1_G$. You know that $h=g^k$ for some $k\ne 0$. What’s the maximum possible number of left cosets that the subgroup $\langle h\rangle$ can have? Can $H$ have any more left cosets than $\langle h\rangle$?

Brian M. Scott
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