Group homology for a metacyclic group

Question

Let $G$ be a finite group, and let $M$ be a finitely generated $G$-module, that is, a finitely generated abelian group on which $G$ acts. We work with the first homology group $$ H_1(G,M).$$

For any cyclic subgroup $C\subseteq G$, we consider the inclusion map $i_C\colon C\hookrightarrow G$ and the induced homomorphism $$ i_{C,*} \colon\, H_1(C,M)\to H_1(G,M).$$

Following Sansuc's paper of 1981, we say that $G$ is metacyclic if all its Sylow subgroups are cyclic. For example, the symmetric group $S_3$ is metacyclic.

Question. Is it true that if $G$ is metacyclic in the sense of Sansuc, then the images $$ {\rm im}\big[i_{C,*} \colon H_1(C,M)\to H_1(G,M)\big]$$ for all cyclic subgroups $C$ of $G$ generate $H_1(G,M)$?

I expect the answer "Yes". Note that when $G$ is cyclic, the answer is obviously "Yes".

Answer 1

The name metacyclic is normally used for a group which is cyclic-by-cyclic (ie. a group $G$ with a cyclic normal subgroup $N$ such that $G/N$ is also cyclic). I will therefore refer to a finite group $G$ with cyclic Sylow subgroups as being Sylow-cyclic. One can show that such groups are in fact metacyclic in the above sense, but I wont use this.

The answer to the OP's question is yes: For $k\geq 1$, $H_k(G,M)$ is a finite abelian group so we can write it as a sum of it’s $p$-torsion parts $H_k(G,M) = \bigoplus_p H_k(G,M)_{(p)}$. If $S$ is a Sylow $p$-subgroup the composite $H_k(G,M)\stackrel{\text{tr}}{\rightarrow} H_k(S,M)\stackrel{i_{S,*}}{\rightarrow} H_k(G,M)$ is multiplication by $[G:S]$. Looking at the $p$-torsion part the composite becomes an isomorphism, so $i_{S,*}\colon H_k(S,M) \rightarrow H_k(G,M)_{(p)}$ is surjective.

Thus $H_k(G,M)$ is always generated by the images corresponding to the Sylow subgroups. If these are all cyclic then $H_k(G,M)$ is generated by images corresponding to cyclic subgroups.

Answer 2

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).