Let $G$ be finite cyclic goup i wont to show that $G$ contain normal subgroup of prime index.

A group G is cyclic if $G$=$ \langle a \rangle$, for some a$\in$$G$.

A finite cyclic group of order n contains a subgroup of order m for each positive integer m which divides n.

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    $\begingroup$ Isn't it clear that the statement you are trying to prove follows immediately from your third sentence? $\endgroup$ – Derek Holt Mar 27 '15 at 20:30
  • $\begingroup$ @DerekHolt the statement that i wont to prove that $G$ contain subgroup of prime index $\endgroup$ – user Mar 27 '15 at 20:33
  • $\begingroup$ And isn't that clear from your third sentence? For example a group of order $100$ contains a subgroup of order $50$ , which has prime index. $\endgroup$ – Derek Holt Mar 27 '15 at 20:36


If $G$ is a cyclic group of order $n$ and there is an $m$ such that $m|n$, then $G$ has a subgroup of order $m$. Furthermore, every subgroup of an abelian group is normal, and every cyclic group is abelian.

Can you see why these facts will get you the result you want? I'll leave their proofs and the finer details to you.

  • $\begingroup$ yes i know this my problem is i could'nt complete this prove $\endgroup$ – user Mar 27 '15 at 20:16
  • $\begingroup$ So if $m|n$, consider the subgroup $H = \langle g^{n/m} \rangle$. What is the order of $H$? For the other results, can you see why cyclic groups are abelian? Can you see why subgroups of abelian groups are normal? For these, go back to the original definitions, and the desired result comes out pretty fast. $\endgroup$ – Kaj Hansen Mar 27 '15 at 20:22
  • $\begingroup$ of course they are clear ,my problem was " subgroup of prime index" but it is ok now thanks $\endgroup$ – user Mar 27 '15 at 20:23
  • $\begingroup$ Ohh, sorry about that; I wasn't sure exactly where you were getting stuck. For a finite group $G$, the index of a subgroup $H \subset G$ is just going to be $|G|/|H|$. So if you want a subgroup $H$ of index $p$, you want $|H| = |G|/p$. $\endgroup$ – Kaj Hansen Mar 27 '15 at 20:26
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    $\begingroup$ yes exactly; thanks for your time. $\endgroup$ – user Mar 27 '15 at 20:28

Hint: If $a$ has order $n$, $a^k$ has order $\dfrac{n}{\gcd(n,k)}$.


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