Suppose $G_1$ and $G_2$ are groups that have the same order. I know that if they are both abelian and cyclic, then they are isomorphic. But does it suffice to only know that they are both cyclic in order to deduce that they are isomorphic ?

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    $\begingroup$ Being cyclic of same order is sufficient. $\endgroup$ – PJK Jun 23 '18 at 15:56
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    $\begingroup$ Cyclic implies abelian. $\endgroup$ – lulu Jun 23 '18 at 16:04

If $G_1$ and $G_2$ are both cyclic of the same order they are of the form: $G_1=<g_1>$ and $G_2=<g_2>$ you can define the morphism (it's gonna be an isomorphism): $$G_1 \rightarrow G_2 \\ g_1^k \mapsto g_2^k$$


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