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A noncentral element of an extraspecial -group has centraliser of index
Statement
Let be an extraspecial -group and let . Then
Facts & Assumptions
Given: An extraspecial -group and an element with .
, the number of left cosets of in (The coset set and the index of a subgroup).
Every conjugacy class of an extraspecial -group whose representative is not central has exactly elements (Every conjugacy class of an extraspecial -group outside the centre has exactly elements).
for a finite group ( is a bijection, so whenever these cardinalities are finite).
For a finite group and , (Lagrange's theorem: for every subgroup of a finite group ).
Proof
The class of has exactly elements, because is not central.
The size of that class is the index of the centraliser of in .
Hence , and Lagrange turns this into .
Remarks
Every centraliser named here is proper, since is noncentral, and maximal in the order sense: index is the smallest index a proper subgroup of a finite -group can have.
Depends on
- Every conjugacy class of an extraspecial $p$-group outside the centre has exactly $p$ elements
- $G/C_G(x)\to\operatorname{Cl}_G(x)$ is a bijection, so $|\operatorname{Cl}_G(x)|=[G:C_G(x)]$ whenever these cardinalities are finite
- The conjugacy class $\operatorname{Cl}_G(x)$ and centralizer $C_G(x)$ of an element
- The coset set $G/H$ and the index $[G:H]$ of a subgroup
- Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$
- The center $Z(G)$ of a group
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
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Sources
- M. van Beek, Topics in Finite p-Groups, Proposition 2.41(ii) (standard reference, not scraped)