# What is an example of a finite centerless group with at least 3 generators?

I need one to test a theory. There are probably many, but I can't seem to think of a single one. My guess is examples are pretty big.

Is there a systematic way to find such examples? Are there databases one can go through to find these things?

-
Are you thinking of finite groups? – Yemon Choi Jan 27 '11 at 21:13
Finite. I'll change it. – Makhalan Duff Jan 27 '11 at 21:15
Can't you take products of simple groups? – José Figueroa-O'Farrill Jan 27 '11 at 21:19
A direct product of simple groups can be 2-generated. – Mark Sapir Jan 27 '11 at 21:38
A direct product of enough copies of a given FSG is not 2-generated. – Jonathan Kiehlmann Jan 27 '11 at 21:48

The group $S_3\wr ({\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_2)$ where $S_3$ is the symmetric group with 6 elements, ${\mathbb Z}_2$ is the group with 2 elements, $\wr$ is the wreath product.

The fact that it does not have a center is proved by inspection. The fact that it needs at least 3 generators follows from the fact that ${\mathbb Z}_2^3$ is its quotient. There are lots of similar examples of course.

Update. In general , if you take any centerless group $G$ and any group $H$ that needs at least 3 generators, then the wreath product $G\wr H$ has both properties (is centerless and needs at least 3 generators). Another way to construct examples is (as Derek Holt comment below shows) to take any centerless finite group $G$ with nontrivial abelianization and take $G\times G\times G$.

-
Isn't that written $S_3\wreath\mathbb Z_2$? – Mariano Suárez-Alvarez Jan 27 '11 at 21:29
$A\wr B$ is the semidirect product of $A^B$ and $B$ where $B$ acts on the direct power $A^B$ by permuting components of vectors. There is a competing notation for the same thing $B\wr A$ used by some dynamics people. But the original (I think due Kaluzhnin and Krasner) is as above. Note that ${\mathbb Z} \wr S_3$ has a center, consisting of constant vectors in the base, also it is not that trivial to show that this wreath product needs 3 generators. – Mark Sapir Jan 27 '11 at 21:35
Why not just $S_3^3$ ? – Derek Holt Jan 27 '11 at 22:54
@Mariano: Yes, $3!=6$. – Mark Sapir Jan 27 '11 at 23:42
@Mariano: Your computation is absolutely correct. – Mark Sapir Jan 28 '11 at 0:50

The smallest example has order 18. It has a normal elementary abelian subgroup $N$ of order 9, and an element $t$ of order 2 such that $txt=x^{-1}$ for all $x \in N$.

-
More generally, an extension of an elementary abelian 3-group $N$ of order $3^r$ by an element of order 2 that inverts everything in $N$ is the smallest example of a centreless group that needs $r+1$ generators. – Derek Holt Jan 28 '11 at 9:15

I can't prove this, but the Rubik's cube group might work.

Update: so this doesn't work. Thanks for the comments.

-
Why do you think so? In any case, gap tells me that the center is cyclic of order $2$ (I'm using the generators from gap-system.org/Doc/Examples/rubik.html) – Mariano Suárez-Alvarez Jan 28 '11 at 8:41
The non trivial central permutation is the one which flips the middle cube of each side. Cute. – Mariano Suárez-Alvarez Jan 28 '11 at 8:45
A quick calculation in GAP or Magma also shows that this group is 2-generated. – Derek Holt Jan 28 '11 at 9:13
@Derek, how did you check? I tried GQuotients(FreeGroup(2), G : findall := false) with G is the cube group module its center (which is centerless) but it seems to take forever---I am pretty sure there are faster ways to do this? – Mariano Suárez-Alvarez Jan 28 '11 at 9:17
@Mariano: gap> G = Subgroup(G, [Random(G), Random(G)] ); true The probability of this returning true seems to be about 0.55. – Derek Holt Jan 28 '11 at 12:07