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I need to show two groups are isomorphic. Since I know they are of the same order, would finding an element that generates the other elements, in both groups, suffice to show that they are isomorphic?

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    $\begingroup$ Yes, that would do it. $\endgroup$ – user99680 Dec 2 '13 at 0:42
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Yes. If you can find an element $x$ which generates the finite group $G$ then it is cyclic. If $G=\langle x\rangle$ and $G'=\langle x'\rangle$ are both cyclic, and $|G|=|G'|$ then $G$ is isomorphic to $G'$ by an isomorphism $\phi\colon G\to G'$ which is defined by $\phi(x^k)=x'^k$. Show that this map is an isomorphism.

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