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enter image description here Can someone justify if these statements are true or false. Just by reading it I can't just fathom a conclusion to these statements. Thanks.

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    Let $G$ be a group, abelian or otherwise. Let $g\in G$ with $g$ not the identity. What can you say about the subgroup *generated by $g$*, $\langle g\rangle = \{e, g, g^2, g^3,\dots\}$? Is it abelian, nonabelian, or unknown? Is it cyclic, not cyclic or unknown? – 2017-01-31
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    It must be cyclic – 2017-01-31
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    What about whether or not it is abelian? – 2017-01-31
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    Obviously if it is cyclic it must be abelian – 2017-01-31
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    Oh yes . I got it . Your above e.g. has answer to both the statements. Got me thinking . Thanks. – 2017-01-31

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If the group is of finite order then that that order is prime or not prime.If it is prime then the group is cyclic hence abelian but your statement says the group is non abelian.So leave this case.Now if the group order is not prime then it has a prime divisor and by Cauchy's theorem there exists an element of that order(divisor).Take that element and generate a subgroup by that element ,the subgroup formed would be cyclic since it is of prime order and hence abelian. This argument satisfies both the statements and if the group is infinite take any non identity element in the group and make a subgroup generated by that element .The subgroup is cyclic always and hence abelian irresepective of whether the group is abelian or not

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    why the downvote – 2017-01-31
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    It wasn't me, but there is no need to invoke Cauchy's theorem here. We aren't looking for a subgroup specifically of prime order, just for any cyclic subgroup with more than one element. It doesn't even specify that they are looking for a proper subgroup for the second part. Just pointing out what properties any subgroup generated by a single element has is enough to complete. – 2017-01-31