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I need help with the following:

Prove that every non cyclic group contains at least $2$ non trivial cyclic groups.

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    How is a cyclic group defined? So what do you know about non-cyclic groups? There's no excuse here for not having provided more context, and including things like what I listed in this comment. – 2017-01-31
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    This question, and the votes to answers of this question, demonstrate just how derailed this site can be. – 2017-01-31
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    @amWhy I totally agree to what you say. – 2017-01-31
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    What's your problem with the question? It's a generic proof in group theory, what context do you need, and why do i have to define what a cyclic group is? Do i also need to define what a non trivial sub group is etc? – 2017-01-31

3 Answers 3

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we know that $|G|>2$ (and even more but it's all you need), so take any element no neutral x. $\langle x \rangle$ can't be G, so take other element in $G\setminus \langle x \rangle$, y. $\langle x \rangle,\langle y \rangle$ are no-trivial cyclic groups.

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Hint: If a group is not cyclic, then it has at least two generators.

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    At least this answer left some work for the OP, who, apparently did nothing on his/her own except to copy and paste the assigned question. Kudos to you! – 2017-01-31
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    You don't know me, or what i did and didn't do on my own. But go on, if rambling out of nowhere makes you happy. – 2017-01-31
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Let $G$ be a non-cyclic group, hence non-trivial: one can pick an element $x \neq 1$. Since $G$ is non-cyclic, $\langle x \rangle \subsetneq G$. So there is an element $y \in G-\langle x \rangle$. Then $\langle y \rangle$ is a second cyclic group.

Note (Feb 1st 2017): you can even stretch the statement as follows: a non-cyclic group has at least three non-trivial cyclic subgroups. And it can be exactly three as $C_2 \times C_2$ shows. How to prove this? See the proof above and consider the subgroup $\langle xy \rangle$. This subgroup clearly differs from $\langle x \rangle$ and $\langle y \rangle$, since $y$ is not a power of $x$ and $x$ is not a power of $y$.