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Prove that all subgroup of a cyclic group generated by $a$ is of the form $\langle a^k\rangle$ where $k|o(G)$

Attempt: Let $o(G)=n$. Let $K=\langle a^k\rangle$. Then $o(a^k)=\frac{n}{gcd(n,k)}$

In general, $k\mid n$ or $k\nmid n$. But how to conclude the remaining. Please help me with simple logic.

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Expanding on the suggestion of Stahl in the comments here is an outline of how to solve the problem:

Let $S$ be a subgroup of a finite group $G=\langle a \rangle$. Then, for each element $s$ in $S$, there exists $n \in \mathbb{N}$ such that $s=a^n$ (why?). Now let $T = \{n \in \mathbb{N}\,|\,a^n \in S, n\geq 1\}$.

(a) Explain why $T$ has a least element $k$.

(b) Explain why $\langle a^k \rangle$ is a subgroup of $S$.

(c) Now suppose that $s \in S$. Then $s=a^n$ for some $n \in \mathbb{N}$.
(i)Use Bezout's identity to show that $a^{\gcd(k,n)}$ is in $S$.
(ii)Explain why this means that $s$ is in $\langle a^k\rangle$.

(d) Use a Corollary of Lagrange's theorem to show that $k$ divides $|G|$.

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  • $\begingroup$ Thanks for the nice expansion. The pointwise style of writing is very good. But one has to solve (a)-(d). I don't know how to prove them all. I know (b) and (d). Please help atleast for (a) and (c) $\endgroup$ – user1942348 Oct 7 '16 at 6:02
  • $\begingroup$ For (a) explain why $T$ is not empty and use the Well-ordering principle $\endgroup$ – Nex Oct 7 '16 at 7:02
  • $\begingroup$ For (c) (i) let $l =\gcd(k,n)$. Bezout's identity implies that there are integers $\alpha$ and $\beta$ such that $l = \alpha k + \beta n$. What does this tell you about $a^l$? $\endgroup$ – Nex Oct 7 '16 at 7:07
  • $\begingroup$ For (c) (ii) Note that $l$ divides $k$ implies $l \leq k$. What does this tell you? $\endgroup$ – Nex Oct 7 '16 at 7:15

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