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.


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 $|G|$.

  • $\begingroup$ 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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