Let $G$ be a finite group, $C(G)$ be a center of $G$, and centralizer of a subset $\{a\}$ (provided $a$ is not an element of $C(G)$) be $C(a)$.

Then, the order of $G$ which satisfies $\#C(G)=2$ and $\#C(a)=\#G/2$ is bounded?

This question is inspired from Probability that $xy = yx$ for random elements in a finite group

If $G$'s order is not bounded, we can say even if we choose two elements from $G$ by not allowing duplication and an identity element, we can say the upper bound of $P(G)$ is $5/8$.

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    $\begingroup$ I'm not sure about your question, but it seems that extraspecial 2-groups satisfy the required conditions. $\endgroup$ – Orat Feb 28 at 7:33
  • $\begingroup$ Thank you very much! Could you tell me why does the second condition is satisfied? $\endgroup$ – buoyant Feb 28 at 9:21
  • $\begingroup$ @DerekHolt That violates the first condition. $\endgroup$ – Orat Feb 28 at 12:22
  • $\begingroup$ Yes you are right! I suspect that that extraspecial 2-groups are the only examples. $\endgroup$ – Derek Holt Feb 28 at 13:17

I understand your question as follows.

Is there a constant $U$ that if $G$ is a finite group with the center $C(G)$ of order two and centralizers $C(a)$ of index two for every $a \in G \setminus C(G)$, then the order of $G$ is bounded above by $U$?

The answer to the above question is negative; and extraspecial $2$-groups provide a counterexample. Since such a group has the center of order two and the order is not bounded, all we have to prove is that the indices of centralizers of non-central elements equal two.

Let $G$ be an extraspecial $2$-group with center $C(G) = \langle t \rangle$ and $a \in G \setminus C(G)$. Because $G' = C(G)$, we have $[a, g] \in \langle t \rangle$ for every element $g \in G$. In other words, a map $$ \phi \colon G \to G', \quad g \mapsto [a, g]$$ is surjective. Hence $G = \phi^{-1}(1) \sqcup \phi^{-1}(t)$. As $\phi^{-1}(1) = C(a)$, we need to investigate what $\phi^{-1}(t)$ is. Let $b \in G \setminus C(a)$. Note that $[a, b] = t$. We claim that $\phi^{-1}(t) = bC(a)$. Suppose $g \in bC(a)$ and write $g = bc$ for some $c \in C(a)$. Then $$[a, g] = [a, bc] = c^{-1}[a, b]c = c^{-1}tc = t.$$ Hence $g \in \phi^{-1}(t)$.

Next, suppose $g \in \phi^{-1}(t)$. Then $$ \begin{align*} [a, b^{-1}g] &= a^{-1}g^{-1}bab^{-1}g = a^{-1}g^{-1}bab^{-1}(a^{-1}a)g\\ &= a^{-1}g^{-1}[b^{-1}, a^{-1}]ag = a^{-1}g^{-1}ag[b^{-1}, a^{-1}]\\ &= [a, g][b^{-1}, a^{-1}] = t[b^{-1}, a^{-1}] = [a, b][b^{-1}, a^{-1}]\\ &= a^{-1}b^{-1}ab[b^{-1}, a^{-1}] = a^{-1}b^{-1}[b^{-1}, a^{-1}]ab = a^{-1}b^{-1}(bab^{-1}a^{-1})ab \\ &= 1. \end{align*} $$ Therefore, $b^{-1}g \in \phi^{-1}(1) = C(a)$.

Now we proved that $G = C(a) \sqcup bC(a)$, so that $C(a)$ has index two.

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    $\begingroup$ Since all centralizers have index $1$ or $2$, for all $g,h \in G$, the elements $g^2$ and $[g,h]$ lie in the intersection of all centralizers, which is $Z(G)$. So $G$ is a $2$-group with $Z(G) = [G,G]$, where $|Z(G)|=2$ and $G/Z(G)$ is elementary abelian i.e. $G$ is an extraspecial 2-group. So these are the only examples. $\endgroup$ – Derek Holt Feb 28 at 14:49

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