# Why if $(\mathbf B, \cdot)$ is a finite order group with prime order then $(\mathbf B, \cdot)$ is cyclic? [duplicate]

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?

Because if $a\in B$, $a\ne1_Β$ and $H=\langle a\rangle=\{a^n\mid n\in\mathbb{Z}\}$, then from Lagrange's Theorem we have that $|H| \mid |B|$ so $a$ has order $p$ and so $H=G$.
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 $b^k.$