In the notes I'm studying from ( again =) ) I read:
If $(\mathbf B, \cdot)$ is a finite order group with prime order then $(\mathbf B, \cdot)$ is cyclic
Could someone give me a justification for this?
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All subgroups of $B$ must be of an order that divides $B.$ If $B$ is of prime order, then its only subgroups are the trivial subgroups.
Suppose $b$ is in $B$ and $b$ is not the identity.
If B has no non-trivial subgroups, it must be the case that everything in B can be expressed as $b^k.$