Let $G$ be a finite solvable group with $Z(G)\neq1$. If for any $1\neq z\in Z(G)$, there exist $g, x\in G$ such that $z=[g,x]\in G^{'}$, can we say that $cl_G(x)=cl_G(zx)$?
Conjugacy classes of a finite solvable group
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abstract-algebra
group-theory
finite-groups
1 Answers
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We can say $cl_G(x)=cl_G(zx)$ and you don't need to assume $G$ is finite or solvable:
The assumptions $z=[g,x]=g^{-1}x^{-1}gx$ and $z\in Z(G)$ give $zxg=xgz=gx$ so $zx=gxg^{-1}$. Hence $x$ and $zx$ are conjugate in $G$ so $cl_G(x)=cl_G(zx)$.