# group presentations that are the most symmetric

for a group G, does there exist a group presentation which is the most "symmetric" by which I mean has the most automorphisms of the group by permuting generators?

• If you look up symmetric presentations of groups you will find some articles that may be of interest to you. – Andreas Caranti Feb 20 '13 at 13:39
• Well, you could give a presentation for $G$ by having the generators be all the (non-trivial) group elements, and the relations being how they multiply together. For example, the Klein $4$-group is $\langle a, b, c; ab=c, ba=c, bc=a, cb=a, ac=b, ca=b, a^2=1, b^2=1, c^2=1\rangle$. Then, all automorphisms permute the generators. – user1729 Feb 20 '13 at 13:40
• @40votes Have done. – user1729 Jul 15 '13 at 9:51

## 1 Answer

Well, you could give a presentation for $G$ by having the generators be all the (non-trivial) group elements, and the relations being how they multiply together. So all relators have length three. For example, the Klein 4-group is as follows. $$\langle a, b, c; ab=c, ba=c, bc=a, cb=a, ac=b, ca=b, a^2=1, b^2=1, c^2=1\rangle$$ Then, all automorphisms permute the generators.

Note that in general not all permutations of the generators will be automorphisms. For example, take $G=C_6$, the cyclic group of order six. Then $G$ has the following presentation. $$G\cong\langle x_1, x_2, x_3, x_4, x_5;$$ $$x_1^2=x_2, x_1x_2=x_3, x_1x_3=x_4, x_1x_4=x_5, x_1x_5=1,$$ $$x_2^2=x_4, x_2x_1=x_3, x_2x_3=x_5, x_2x_4=1, x_2x_5=x_1,$$ $$x_3x_1=x_4, x_3x_2=x_5, x_3^2=1, x_3x_4=x_1, x_3x_5=x_2,$$ $$x_4x_1=x_5, x_4x_2=1, x_4x_3=x_1, x_4^2=x_2, x_4x_5=x_3,$$ $$x_5x_1=1, x_5x_2=x_1, x_5x_3=x_2, x_5x_4=x_3, x_5^2=x_4\rangle$$ Note that $x_1$ has order six and $x_2$ has order three. Then any permutation which switches $x_1$ and $x_2$ is not an automorphism of $G$.

More generally, suppose $G$ is finitely presented and has finite automorphism group (we are trying to generalise the finite case, so an infinite example is $\mathbb{Z}$, which has two automorphisms, the trivial one and the one which switches $1$ and $-1$). Then $G$ has a finite presentation such that every automorphism permutes all the generators, for example $\mathbb{Z}$ can be given by a presentation $\langle a, b; ab=1\rangle$ and the non-trivial automorphism is simply $a\leftrightarrow b$.