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{${G,*}$} is a group of order 15 with identity element ${e}$. There is an element ${a\in G}$ such that ${a^3\neq e}$ and ${a^5\neq e}$. Prove that {${G,*}$} is a cyclic group with generator ${a}$.

This is what I have written: By Lagrange's theorem subgroups can only exist of order 1,3,5 or 15. Since ${a\neq e}$, ${a^3\neq e}$ and ${a^5\neq e}$ then ${a^{15}=e}$. Therefore the order of ${a}$ is the same as the order of ${G}$. Therefore ${G=<a>}$, and hence G is cyclic.

Is this correct? Have I missed anything?

RedG
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  • Pretty good I would say :) Indeed since the period of $a$ is $15$ it will necessarily generate all members of the group when "iterated". Otherwise it would have had a smaller period which is false. – Zubzub Jul 29 '16 at 13:43
  • The logic employed is a special case of the frequently useful Order Test. – Bill Dubuque Jul 29 '16 at 17:12

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It is perfect.

Perhaps you skipped a few steps, like the fact that the order of $a$ is the order of the subgroup $\langle a \rangle$, which must be $1,3,5$ or $15$ by lagrange's theorem.

And then noticing that if the order is $1$ or $3$ then $a^3=e$ and if the order is $5$ then $a^5=e$. So you finally conclude the order is $15$.

Asinomás
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