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Everything follows from the p-primary decomposition theorem, which is nicely explained in Ken Brown's group cohomology bible (Kasper's answer is the restriction-corestriction argument written circa Theorem III.10.3 for which you see that the p-primary component is the set of invariant-elements of the cohomology of a p-Sylow subgroup). As an aside / corollary, your group has periodic cohomology (Theorem VI.9.5), so you can learn a lot about your group / cohomology.

And Sansuc's definition of metacylic is indeed unorthodox -- the usual definition is that it has an extension of a cyclic group by a cyclic group. In particular, a group whose Sylow subgroups are cyclic is necessarily metacyclic, but the converse is not true: $\mathbb Z/3\mathbb Z\times S_3$ has cyclic commutator subgroup (of order 3) whose quotient is cyclic (of order 6), but $\mathbb Z/3\mathbb Z\times\mathbb Z/3\mathbb Z$ is Sylow. What I think is true of a metayclic group is that its p-Sylow subgroups are cyclic for p the smallest prime divisor of |G|.

Everything follows from the p-primary decomposition theorem, which is nicely explained in Ken Brown's group cohomology bible (Kasper's answer is the restriction-corestriction argument written circa Theorem III.10.3 for which you see that the p-primary component is the set of invariant-elements of the cohomology of a p-Sylow subgroup). As an aside / corollary, your group has periodic cohomology (Theorem VI.9.5), so you can learn a lot about your group / cohomology.

And Sansuc's definition of metacyclic is indeed unorthodox -- the usual definition is that it has an extension of a cyclic group by a cyclic group.* In particular, a group whose Sylow subgroups are cyclic is necessarily metacyclic, but the converse is not true: $\mathbb Z/3\times S_3$ has cyclic commutator subgroup (of order 3) whose quotient is cyclic (of order 6), but $\mathbb Z/3\times\mathbb Z/3$ is Sylow. What I think might be true of a metacyclic group is that its $p$-Sylow subgroups are cyclic for $p$ the smallest prime divisor of $|G|\ne p^n$.

*OK there is a 3rd notion of metacyclic, that its commutator subgroup and corresponding quotient are cyclic ($Q_8$ is not metacyclic in this sense but it is metacyclic in the cyclic-by-cyclic extension sense).

Just to tack onto Kasper's response. Everything follows from the p-primary decomposition theorem, which is nicely explained in Ken Brown's group cohomology bible. In particular, such a group has periodic cohomology (Theorem VI.9.5). So you can learn a lot about your group / cohomology.

And unless I've misunderstand the literature and proofs, Sansuc's definition of metacylic is unorthodox (the usual definition is that it has an extension of a cyclic group by a cyclic group). In particular, a group whose Sylow subgroups are cyclic is necessarily metacyclic, but the converse is not true: $\mathbb Z/3\mathbb Z\times S_3$ has cyclic commutator subgroup (of order 3) whose quotient is cyclic (of order 6), but $\mathbb Z/3\mathbb Z\times\mathbb Z/3\mathbb Z$ is Sylow. What I think is true of a metayclic group is that its p-Sylow subgroups are cyclic for p the smallest prime divisor of |G|.

Everything follows from the p-primary decomposition theorem, which is nicely explained in Ken Brown's group cohomology bible (Kasper's answer is the restriction-corestriction argument written circa Theorem III.10.3 for which you see that the p-primary component is the set of invariant-elements of the cohomology of a p-Sylow subgroup). As an aside / corollary, your group has periodic cohomology (Theorem VI.9.5), so you can learn a lot about your group / cohomology.

And Sansuc's definition of metacylic is indeed unorthodox -- the usual definition is that it has an extension of a cyclic group by a cyclic group. In particular, a group whose Sylow subgroups are cyclic is necessarily metacyclic, but the converse is not true: $\mathbb Z/3\mathbb Z\times S_3$ has cyclic commutator subgroup (of order 3) whose quotient is cyclic (of order 6), but $\mathbb Z/3\mathbb Z\times\mathbb Z/3\mathbb Z$ is Sylow. What I think is true of a metayclic group is that its p-Sylow subgroups are cyclic for p the smallest prime divisor of |G|.

Just to tack onto Kasper's response. Everything follows from the p-primary decomposition theorem, which is nicely explained in Ken Brown's group cohomology bible. In particular, such a group has periodic cohomology (Theorem VI.9.5). So you can learn a lot about your group / cohomology.

And unless I've misunderstand the literature and proofs, Sansuc's definition of metacylic is unorthodox (the usual definition is what Kasper wrote). In particular, a group whose Sylow subgroups are cyclic is necessarily metacyclic, but the converse is not true: $\mathbb Z/3\mathbb Z\times S_3$ has cyclic commutator subgroup (of order 3) whose quotient is cyclic (of order 6), but $\mathbb Z/3\mathbb Z\times\mathbb Z/3\mathbb Z$ is Sylow. What I think is true of a metayclic group is that its p-Sylow subgroups are cyclic for p the smallest prime divisor of |G|.

Just to tack onto Kasper's response. Everything follows from the p-primary decomposition theorem, which is nicely explained in Ken Brown's group cohomology bible. In particular, such a group has periodic cohomology (Theorem VI.9.5). So you can learn a lot about your group / cohomology.

And unless I've misunderstand the literature and proofs, Sansuc's definition of metacylic is unorthodox (the usual definition is that it has an extension of a cyclic group by a cyclic group). In particular, a group whose Sylow subgroups are cyclic is necessarily metacyclic, but the converse is not true: $\mathbb Z/3\mathbb Z\times S_3$ has cyclic commutator subgroup (of order 3) whose quotient is cyclic (of order 6), but $\mathbb Z/3\mathbb Z\times\mathbb Z/3\mathbb Z$ is Sylow. What I think is true of a metayclic group is that its p-Sylow subgroups are cyclic for p the smallest prime divisor of |G|.