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Prop: If $G$ is a finite simple group, then a maximal subgroup of $G$ is trivial or has composite order

Proof: A maximal subgroup of $G$ being trivial clearly corresponds to $G$ being cyclic of prime order. Assume, then, that $G$ is non-abelian.

If $G$ has a maximal subgroup $C$ of prime order, then the action of $G$ on cosets of $C$ is Frobenius. Thus $G$ is a Frobenius group, $C$ is a Frobenius complement and $G$ contains a Frobenius kernel, i.e. $G$ is not simple, and . QED

So this answers your first questionis answered. As for your more general question about finite groups: There is loads . Well, again, if a group has a maximal subgroup of stuff in the literature about prime order, then it is Frobeniusgroups, so you should do some reading! consult the literature on Frobenius groups. For this I particularly recommend Isaac's "Finite Group Theory" and Passman's "Permutation groups".

Examples of groups with a maximal subgroup of prime order include dihedral groups of order $2m$ ($m$ odd) or, more generally $C_n \rtimes C_p$ where $p$ is a prime and $C_p$ acts semi-regularly on $C_n$.

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If a group $G$ has a maximal subgroup $C$ of prime order, then the action of $G$ on cosets of $C$ is Frobenius. Thus $G$ is a Frobenius group, $C$ is a Frobenius complement and $G$ contains a Frobenius kernel, i.e. $G$ is not simple, and your first question is answered.

As for your more general question about finite groups: There is loads of stuff in the literature about Frobenius groups, so you should do some reading! I particularly recommend Isaac's "Finite Group Theory" and Passman's "Permutation groups".

Examples of groups with a maximal subgroup of prime order include dihedral groups of order $2m$ ($m$ odd) or, more generally $C_n \rtimes C_p$ where $p$ is a prime and $C_p$ acts semi-regularly on $C_n$.