# How “abelian” can a non-abelian group be?

Something I have been wondering: in general, is there a bound for how many elements in a finite non-abelian group $G$ can commute with every other element? Equivalently, is there is a bound for the order of the center relative to the order of the group?

And, if $G$ is non-abelian and infinite, does this question make any sense? (ie does it make sense to consider the ratio $\frac{|Z(G)|}{|G|}$?)

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An obvious remark: Based on Lagrange's theorem, the center couldn't contain more than half the elements of the group. There is probably a better estimate than this though. –  jmracek Nov 13 '12 at 15:28
I think there is some result along the lines that the proportion of pairs of elements $(g_1,g_2)$ with $g_1,g_2 \in G$ and $g_1g_2=g_2g_1$ is at most 5/8 (?) in any nonabelian group $G$. –  Derek Holt Nov 13 '12 at 17:08
@DerekHolt: Yes I think that's correct, and there a lot of nice results on the "commutativity degree" of a finite group, for example this thesis: math.wfu.edu/publications/Student/anna_castelaz.pdf . –  user641 Nov 14 '12 at 0:00
@Steve, what a great reference! I might add that the commutativity degree is of course invariant under isomorphism, but even invariant under isoclinism. This was proved by A. Erfanian and F. Russo in 2008 I believe, see bit.ly/T5dkOj for example. –  Nicky Hekster Nov 18 '12 at 21:31

Although what Hagen von Eitzen says is true, there is a much stronger restriction. The index of the center of a nonabelian group can never be a prime. So if the center is of index 2, then that group is actually abelian and so the center is actually the whole group.

Let's see why this is true.

Claim: The center $Z(G)$ of a group $G$ can never have index $p$ in $G$ for any prime $p$.

proof: As we so often do, suppose not; i.e. suppose we have a group $G$ with center $Z(G)$ of index $p$. Then $Z(G)$ is normal, so we can consider the quotient group $G/Z(G)$, which is a group of size $p$. Thus it is cyclic, say generated by an element $g$, and abelian.

In particular, this means that every element of $G$ can be written as $g^iz$ for some element $z$ in the center. So take two elements $a = g^iz_1$ and $b = g^jz_2$ in $G$, and consider their product.

$$ab = g_iz_1g^jz_2 = g^ig^jz_1z_2 = g^jg^iz_2z_1=g^jz_2g^iz_1 =ba$$

Where I used that the $g^i$ commute, and elements in the center commute with everything. Thus this shows that the group is abelian, contrary to the fact that the center was not the whole group. Contradiction. $\diamondsuit$

So the center cannot contain more than half the elements of the group, and in fact can't contain exactly half the elements of the group. But there is no reason why the index of the center couldn't be $6$ infinitely often, visibly by taking semidirect products of the symmetric group on 3 symbols with larger and larger abelian groups.

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I was just about to post that it is impossible for the centre to have prime index, because if $G/Z$ is cyclic, $G$ is abelian. So the largest possible centre has order $|G|/4$. Note: The proof applies to all cases where $G/Z$ is cyclic. –  Mark Bennet Nov 13 '12 at 15:45
Does $H$ and $G/H$ being abelian impliy $G$ abelian? If yes, then $|G:Z| \geq 6$ as $S_3$ is the smallest nonabelian group. Klein 4 is abelian, isn't it? –  Karolis Juodelė Nov 13 '12 at 15:48
@Karolis: Thanks for that - I changed it to $S_3$. You're exactly right - I must have been secretly equating cyclic and abelian when I wrote that. –  mixedmath Nov 13 '12 at 15:56
@KarolisJuodelė: The quaternions are a counterexample. The center is $H=\mathbb Z_2, \ G/H=V_4$, the Klein 4-group, which is abelilan. –  Ross Millikan Nov 13 '12 at 16:19
Thanks Ross - you're right-er. –  mixedmath Nov 13 '12 at 18:12

Note that if $A,B$ are two groups then $Z(A \times B) = Z(A) \times Z(B)$. This implies that the size of the center of a non-abelian group can be arbitrarily large (take $A$ an abelian group of a desired size and $B$ a non-abelian group with trivial center, for example any non-abelian simple group), and still the bound $|Z(G)| \leq |G|/4$ (deriving from the fact that if $G/Z(G)$ is cyclic then $G$ is abelian) cannot be improved (consider e.g. the group of quaternions, $Q_8$). So, both $|G|$ and $|Z(G)|$ can be arbitrarily large (even infinite), and hence you don't have much control on that. I guess you could ask about specific numbers instead. For example, your question could be the following:

(1) what are the numbers $n$ for which there exists a group $G$ with $Z(G)=\{1\}$ and $|G|=n$ ?

I believe this is a very non-trivial question. For instance, it is known that a number $n$ is "cyclic", i.e. such that every group of order $n$ is cyclic, if and only if $(n,\varphi(n))=1$. And there is a similar result concerning nilpotent groups (which always have non-trivial center), see here. My guess would be that the answer to (1) is: the non-nilpotent numbers. But I don't know.

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It is not true that non-nilpotent numbers characterize orders of groups with trivial center. For example, every group of order $28$ has nontrivial center but there exists a non-nilpotent group of order $28$. Another example is $40$. –  Mikko Korhonen Nov 13 '12 at 17:02
You can consider the index of $Z(G)$ in $G$, which may be finite even if both $G$ and $Z(G)$ are infinite. Other than that, if $G$ is not abelian, that index is at least $2$, i.e. the center contains at most half the group elements.
I get that LaGrange implies that the order of any proper subgroup is at most $|G|/2$, but I guess my question is, can we keep on increasing the order of the unknown $G$ and keep on finding groups with centers of index 2? –  tacos_tacos_tacos Nov 13 '12 at 15:32
The center cannot have index 2 if the size of your group is larger than 2. Do you mean order 2? If $X$ is a group with trivial center (for example, a non-abelian simple group) then $X \times C_2$ has center of order 2. Note that the bound $|Z(G)| \leq |G|/4$ is best possible (take e.g. $Q_8$). –  Martino Nov 13 '12 at 15:48