In a group $G,$ the centralizer of $a \in G$ is the subgroup $$C(a)=\{g \mid ga=ag\}.$$

In a non-abelian group with $|G|=n$ and center $Z$ we have for each $a \notin Z$ that $Z \subsetneqq C(a) \subsetneqq G.$ This shows that $|Z| \le \frac{n}{4}$ and that, when $|Z|=\frac{n}{4},$ each $a \notin Z$ commutes with exactly half the elements so $$\sum_{a \notin Z}|C(a)|=\frac{3n^2}8.$$

My question is if the converse is true:

  • If $\sum_{a \notin Z}|C(a)| = \frac{3n^2}8$ does it follow that $|Z|=\frac{n}{4}?$
  • Can it happen that $\sum_{a \notin Z}|C(a)| \gt \frac{3n^2}8?$

So really that is the converse of $$ |Z|=\frac{n}4 \text{ implies } \sum_{a \notin Z}|C(a)| \ge \frac{3n^2}8. $$

A near miss is the semi-dihedral group SD16 which has size $16$ and center of size $2$. It has a cyclic subgroup of size $8$ whose $6$ non-central members have $|C(a)|=8$ and the other $8$ elements have order $4$ and $|C(a)|=4$ for a total of

$$\sum_{a \notin Z}|C(a)|=6\cdot 8+8 \cdot 4=\frac{5n^2}{16}.$$

A group with $|Z| \lt \frac{n}4$ and all other elements having $|C(a)|=\frac{n}{2}$ would provide an example with $\sum_{a \notin Z}|C(a)|=\frac{(n-|Z|)n}{2} \gt \frac{3n^2}8.$

  • $\begingroup$ Another example of a group achieving $5n^2/16$ is the subgroup of affine transforms of $F_4^2$ generated by all translations and $(x,y)\mapsto (y,x)$. Perhaps, it can be derived from your example... Also, notice that $S_3$ gets $n^2/3$. $\endgroup$ – Ilya Bogdanov Sep 8 '16 at 10:30
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    $\begingroup$ I will add more details later, but a group G with your property has order divisible by 8, and has all its odd order Sylow subgroups normal. Also, for at most one prime p can G have a non-Abelian Sylow p-subgroup, and p can only be 2 or 3. $\endgroup$ – Geoff Robinson Sep 8 '16 at 11:54
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    $\begingroup$ It is also the case that the derived group [G,G] has order at most 5 ( this and the previously mentioned properties can all be deduced from Lemma 2 of the paper that Bob Guralnick and I wrote on the Commuting Probability in Finite Groups). $\endgroup$ – Geoff Robinson Sep 8 '16 at 15:23
  • $\begingroup$ I guess for the second question one can take the direct product of several copies of $D_8.$ $\endgroup$ – Aaron Meyerowitz Sep 8 '16 at 17:14
  • $\begingroup$ I don't think this direct product works. A typical element in this product has only about half central coordinates, so its centralizer is relatively small. Concretely, if you take the direct product of $n$ copies of $D_8$, the sum equals $(8+4)^n=12^n$. $\endgroup$ – Ilya Bogdanov Sep 9 '16 at 6:36

Firstly, any finite non-Abelian $2$-group $G$ as in the last remark of the question has centre of index $4$. Suppose otherwise and set $Z= Z(G).$ Suppose that each element of $G \backslash Z$ has centralizer of index $2$. If $G$ has two different Abelian maximal subgroups $M$ and $N$ then $M \cap N = Z$ and has index $4$. If there is a unique Abelian maximal subgroup $M$ of $G$, then for any $x$ outside $M$, we see that $<x> C_{M}(x)$ is Abelian, is different from $M$ and has index $2$ in $G$, a contradiction. Hence we may suppose that $G$ has no Abelian maximal subgroup.

By induction, we may suppose that $[M:Z(M)] = 4$ for each maximal subgroup $M$ of $G$, so every Maximal subgroup of any such $M$ is Abelian. But as above , we then see that for any $x$ outside $M$, the centralizer of $x$ in $G$ is an Abelian maximal subgroup of $G$, contrary to assumption. Hence we do have $[G:Z] = 4, $ as claimed. Note thatvweay suppose that $Z \leq M, $ otherwise $G = ZM,$ and then $ZX$ is a central subgroup of $G$ of index $4,$ where $X = Z(M).$

Here is a sketch proof that a group $G$ satisfying the first condition must be a $2$-group- well, strictly a direct product of a 2-group and an Abelian group of odd order. You want $G$ to have $|Z|+ 3|G|/8$ conjugacy classes, where $Z=Z(G)$. If we choose $G$ of minimal order subect to having this property, then $G$ can not be expressed in the form $A \times H$ for $A,H$ proper with $A$ Abelian, so suppose that this is the case.

Suppose now that $G$ is not a $2$-group. Then $G = SN$ where $S$ is a $2$-group and $N$ is a non-trivial normal $2$-complement. If $[S,N]= 1,$ then $G=S \times N,$ contrary to hypothesis. Hence $[S,N]$ has order $3$ or $5$, in which case $S$ is Abelian as $[G,G]$ has order at most $5$. It follows that $N$ is either a $3$-group or a $5$-group. Also $N$ must be Abelian, since we can't have $[S,N] \leq [N,N]$ by properties of coprime automorphisms.

Now $G = (S[N,S]) \times C_{N}(S)$ so $C_{N}(S) = 1$ and $Z = Z(G)$ is a $2$-group, and now $N = [N,S]$ has order $3$ or $5$.

Suppose that $|N| = 5.$ Now $G$ has $|S|$ linear characters and $\frac{15|S|}{8}+ |Z|- |S|$ non-linear irreducible characters, so that $5|S| \geq |S| + d^{2}(\frac{7|S|}{8}+ |Z|)$, where $d$ is the smallest degree of a non-linear irreducible character of $G$, which is easily seen to force $d=2.$

This in turn forces $[S:Z]= 2$, which easily yields a contradiction.

Hence in the case under consideration, we must have $|N| = 3,$ and then again $[S:Z]=2.$ But now, each non-linear irreducible character of $G$ has degree $2$ by a Theorem of Ito, and we have $3|S| = |S| + 4(\frac{9|S|}{8} -\frac{|S|}{2}),$ which is a contradiction.

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