Given a group $G$ and $a \in G$, prove that the set $\langle a\rangle=\{a^k\}$ is a subgroup of $G$.
I am not able to see that why will the inverses be present.
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Of course it's a subgroup because it's a cyclic group . – Gankedbymom Apr 25 '17 at 13:01
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By definition, $S=\langle a\rangle$ is the subgroup of $G$ generated by an element $a\in G$, see here. It consists of the elements $\{a^k\mid k\in \mathbb{Z}\}$. If $a$ has finite order $n$, then $S=\{1,a,a^2,\ldots ,a^{n-1}\}$. Then the inverse is given by $(a^k)^{-1}=a^{-k}=a^{n-k}$.
Dietrich Burde
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