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What are the automorphism groups of non-abelian groups of order 27? (there are two non-abelian groups of order 27).

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Could you please provide some motivation? –  Alex B. Jan 22 '11 at 4:19
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If G is a group of order p^3.q (p=/=2, q=/=3), then G always has a normal Sylow subgroup! So all groups of this type can be obtained using semidirect products (which are easy, can write presentations explicitely, and easier than other extention types, as cyclic, central etc). But to get all possible semidirect products, we would like to know automorphisms of groups of order p^3. I am working the case when G has order 54=2.(3^3). Here Sylow-3 subgroup is normal, order is 27, what are their automorphism groups? (for abelian case its easy). –  user8186 Jan 22 '11 at 5:00

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The non-abelian group of order $p^3$ with no elements of order $p^2$ is the Sylow $p$-subgroup of $\operatorname{GL}(3,p)$. Its automorphism group can also be viewed as a group of $3\times3$ matrices, the affine general linear group, $$\operatorname{AGL}(2,p) = \left\{ \begin{pmatrix}a & b& e\\ c& d& f\\ 0 & 0 & 1\end{pmatrix} : a,b,c,d,e,f \in \mathbb{Z}/p\mathbb{Z},\; ad-bc ≠ 0 \right\}, $$ which is the semi-direct product of $\operatorname{GL}(2,p)$ on its natural module.

This description is reasonably famous, especially when considering non-abelian groups of order $p^{2n+1}$ with no elements of order $p^2$ whose center and derived subgroup have order $p$. Instead of $\operatorname{GL}(2,p)$ you get a variation on $\operatorname{Sp}(2n,p)$, that simplifies to $\operatorname{GL}(2,p)$ when $n=1$.

The non-abelian group of order $p^3$ with an element of order $p^2$ and $p ≥ 3$ has as its automorphism group a semi-direct product of $\operatorname{AGL}(1,p)$ with the dual of its natural module, so you get all $3×3$ matrices $$\left\{ \begin{pmatrix}a & b& 0\\ 0& 1& 0\\ c & d & 1\end{pmatrix} : a,b,c,d \in \mathbb{Z}/p\mathbb{Z},\; a ≠ 0 \right\}. $$

In both cases the "module part" of the semi-direct product is the group of inner automorphisms and the quotient ( $\operatorname{GL}(2,p)$ and $\operatorname{AGL}(1,p)$ ) are the outer automorphism groups.

You can read about some of this in section A.20 of Doerk–Hawkes, or Winter (1972).

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How the structure of permutation groups AGL(1,p)=Z_p:Z_(p-1)? –  user9653 Apr 16 '11 at 8:04
    
AGL(1,p) = { [a,b;0,1] : a,b in Z/pZ, a ≠ 0 } has a normal subgroup ( "a=1" ) of order p with complement ( "b=0" ) of order p-1. –  Jack Schmidt Apr 16 '11 at 13:04

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