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".
Best Answer
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.