# Is the order of this group bounded?

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 $$5/8$$.

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

• 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. – Derek Holt Feb 28 at 14:49