Say that a group $Q$ is a quasiquaternion if $Q=CU$, where $C$ and $U$ are cyclic subgroups of $Q$, $C$ is normal and $C \cap U = \mathbf{Z}(Q)$, where $\mathbf{Z}(Q)$ is a center of $Q$.
Let $G$ be a group and $G/Z$ is a quasiquaternion, where $Z \subseteq G'\cap\mathbf{Z}(G)$, where $G'$ is a derived subgroup. Show that $Z=1$ (that is $Z$ is trivial.)
I have shown that
- $G=C\cdot U$, where $C,U$ are abelian subgroups, $U/Z$ and $C/Z$ are cyclic, $C$ is normal, $Z\subset U\cap C$.
- $\mathbf{Z}(G)=U\cap C$;
- $G' \cong C/\mathbf{Z}(G)$. And hence $G'$ is cyclic.
But I dont know how to prove that $Z$ is trivial. I note that it is not true that $G'\cap \mathbf{Z}(G)$ is trivial (for example $G:=Q_8$).
I will be gratefull for hints and ideas.
You have used $C$ and $U$ twice in your post with different meanings, which is confusing! I will replace the second usage by $A,B$. So now $G=AB$ with $A/Z=C$ and $B/Z = U$.
You have shown that $Z(G) = A \cap B$, and $(A \cap B)/Z = C \cap U = Z(Q)$, so $|Z(G)| = |Z||Z(Q)|$.
You have also shown that $|G'| = |A/Z(G)| = |C/Z(Q)|$ (using Lemma 4.6 of Isaacs). By the same argument applied to $Q$, we gave $|Q'| = |C/Z(Q)|$, so $|G'|=|Q'|$.
But since $Z \le G'$, we have $|G'| = |Z||Q'|$, so $Z=1$.
That was assuming that $Q$ is finite. (Note that finite groups have finite Schur Multipliers, so $Q$ finite implies $Z$ finite.) If $C$ is finite, then $U$ cannot be infinite, because that would imply $U \cap Z(Q)$ infinite. If $C$ is infinite, then ${\rm Aut}(C) \cong C_2$, so the only possibility for $Q$ is the infinite dihedral group with $Z(Q) = 1$. But that has trivial Schur Multiplier, so the result still holds.