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This is just something I conjectured after playing around with some examples of groups.

Prove that a subgroup of a cyclic group is cyclic.

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It’s true, and it’s a basic, standard result in elementary group theory. – Brian M. Scott Apr 10 '12 at 4:01
If it's just something you're wondering idly about, then formulating it as if it were a homework exercise for the readership ("Prove that ...") is not really polite. – Henning Makholm Apr 10 '12 at 9:33
@justwatching Fun to know: if $G$ is a group of which all proper subgroups are cyclic, then $G$ itself does not have to be cyclic. Can you give an example? Also an example where $G$ is infinite? – Nicky Hekster Apr 10 '12 at 14:33
up vote 1 down vote accepted

Let $G=\langle a \rangle$ be cyclic group generated by 'a'& H be a subgroup of group G let m be the smallest positive integer such that $a^m \in H$ then clearly $\langle (a^m)^q \rangle\ \in H$ $\langle a^m\rangle\subset H$

let $a^t\in H$

applying division algorithm for integer t&m $\exists q,r\in Z$ such that

$t=m.q+r$, where $0\leq r <m $

$\implies r =t-mq$

$\implies a^r=a^t.a^{-mq}$

$\implies a^t.(a^m)^{-q}\in H$

$\implies a^r \in H$ , r< m

If $r\neq 0$ then r is the smallest positive integer such that $a^r\in H$

Which is a contradiction {$\because$ m is the smallest positive integer }

\therefore $r=0 $

$\implies t=m.q $

$\implies a^t=a^m.q$

$\implies a^m.q \in \langle a^m \rangle$

$\implies H \subset \langle a^m \rangle$

$\therefore H=\langle a^m \rangle$

Thus, Subgroup of cyclic group is cyclic

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Hint. Let $G=\langle g \rangle$ be a cyclic group. Let $H$ be a subgroup of $G$. If $H$ is trivial, then $H=\langle e\rangle$. Otherwise, prove that $H=\langle g^k\rangle$, where $k$ is the smallest positive integer such that $g^k\in H$.

You'll have to prove that if $H\neq\{e\}$, then there is such a $k$, and that the subgroup it generates is equal to $H$.

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What does the notation $\langle g \rangle$ mean? – justwatching Apr 10 '12 at 4:15
<g> is the subgroup generated by the element g – yoyostein Apr 10 '12 at 4:28

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