Prove/Disprove:
Every subgroup generated by one element is cyclic
If $o(a)=n$ so $a^{-1}=a^{n-1}$
By definition a group is cyclic if for an element $a\in G$ , $\langle a \rangle=G$ (it can be one or more elements, in particular one element)
By definition $a^{0(a)}=e$ because $a\in G$ is a group there is $a^{-1}\in G$ and $G$ is closed under the operation So $$a^{0(a)}=e\iff a^{n}=e$$ Multiply both sides by $a^{-1}$
$$a^{n}\cdot a^{-1}=e\cdot a^{-1}\iff a^{n-1}=a^{-1}$$
Are the proofs valid?
Yes, you're correct.
The
bit is somewhat unnecessary, however.
It could be more rigorous if you referenced the lemma that is the law of indices you used at the end too. That's somewhat pedantic though.