In an article by Bob Oliver "Simple fusion systems over p-groups with abelian subgroup of index p" in Notation 2.2., he defines a set $Z_2 = Z_2(S)$ where $S$ is a nonabelian $p$-group. The problem is that I don't know what $Z_2(S)$ means. I know $Z(S)$ is the center, but I have never seen $Z_2(S)$. Can somebody help me?
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$\begingroup$ Perhaps a group-ring is meant? A bit like polynomials, but with coefficients drawn from $\mathbb{Z}_2$ and exponents from $S$? $\endgroup$– hardmathJun 7, 2014 at 14:27
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4$\begingroup$ Maybe it is the second center, Z(G/Z). $\endgroup$– i. m. soloveichikJun 7, 2014 at 14:29
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$\begingroup$ Second center would make sense. Because later he talks about $Z_2/Z$ where $Z= Z(S)$, so $Z(S)$ is a subgroup of $Z_2(S)$. $\endgroup$– user129954Jun 7, 2014 at 14:33
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1 Answer
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As i.m. soloveichik points out in the comments, $Z_2(G)$ presumably refers to the second center of the group $G$, i.e. the second term in the upper central series.
This is indeed a subgroup of $G$, and it has several equivalent definitions:
It is the inverse image of $Z\bigl(G/Z(G)\bigr)$ under the quotient homomorphism $G \to G/Z(G)$.
It is the set $\{g\in G \mid [g,h]\in Z(G)\text{ for all }h\in G\}$.
It is the set $\{g\in G \mid [[g,h],k]=1\text{ for all }h,k\in G\}$.
Note that $Z_2(G)$ is a normal (and indeed characteristic) subgroup of $G$, and that $Z_2(G)$ always contains $Z(G)$. Moreover, $Z_2(G)$ is equal to $G$ if and only if $G$ is nilpotent of class 2, i.e. if and only if $G$ is a central extension of an abelian group.
