If we know that $G'$ is a subgroup of a $Z(G)$, can we conclude that $G'=Z(G)$ ?
What I can see is that $G/Z(G)$ is abelain and it can not be cyclic but no further ..
If we know that $G'$ is a subgroup of a $Z(G)$, can we conclude that $G'=Z(G)$ ?
What I can see is that $G/Z(G)$ is abelain and it can not be cyclic but no further ..
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No not true, take $G$ non-trivial abelian, then $G'=1$ and $Z(G)=G$. For an non-abelian example, take $G=A \times Q$, where $A$ is non-trivial abelian and $Q$ is the quaternion group of order $8$. Note $Z(Q)=Q'$.