# Automorphisms of a non-abelian group of order $p^{3}$

In the book "Structure of Groups of Prime Power order (Charles Richard Leedham-Green, Susan McKay), there is an exercise (2.1.10), which asks to show that the automorphism group of ($\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_p$ is ($\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes GL(2,p)$. To prove this, first, it is easy to show that there is a short exact sequence $1 \rightarrow \mathbb{Z}_p \times \mathbb{Z}_p \rightarrow Aut((\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_p) \rightarrow GL(2,p) \rightarrow 1$. Therefore it remains to show that "this exact sequence is split". How to show it?

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@Rahul: Could you typeset your question and title? – user17762 Jan 28 '11 at 3:13

Here is a new, less coordinated version that motivates some of the correct coordinate changes.

Let V be a vector space, let W = V∧V be the product, and let G be the wedgey group on V⊕W with multiplication (v1,w1)⋅(v2,w2) = (v1+v2, w1+w2 + v1∧v2). Powers obey the rule (v,w)n = (nv,nw). Commutators obey the rule [(v1,w1),(v2,w2)] = [0,2(v1∧v2)]. In particular, G always has nilpotency class 2 when 0≠2 (and is abelian otherwise), and has exponent p whenever V is defined over a field of characteristic p. When V is one-dimensional over the field Z/pZ of p elements, G is the unique non-abelian group of order p3 and exponent p (the group in question).

Consider a block matrix of the form $$M = \begin{bmatrix} A & b \\ . & \Lambda^2(A) \end{bmatrix}$$ where A in End(V) and b in Hom(V,W), with the formal multiplication (v,w)⋅M = ( vA, vb + w⋅Λ2(A) ). When V is 2-dimensional, W is 1-dimensional and Λ2(A) is just multiplication by det(A). In general, (v1∧v2)⋅Λ2(A) = (v1⋅A) ∧ (v2⋅A). When dim(V)=2, then every element of Hom(V,W) can be written as b = ∧u which takes v in V to v∧u in W.

This matrix M defines a homomorphism of G:

$$\begin{array}{l} (v_1, w_1)M \cdot (v_2, w_2)M \\ = (v_1\cdot A, ~w_1\cdot\Lambda^2(A) + v_1\cdot b) \cdot (v_2\cdot A, ~w_2\cdot\Lambda^2(A) + v_2 \cdot b) \\ = (v_1\cdot A + v_2\cdot A, ~w_1\cdot\Lambda^2(A) + v_1 \cdot b + w_2\cdot\Lambda^2(A) + v_2 \cdot b + (v_1\cdot A)\wedge(v_2\cdot A) ) \\ = ( (v_1+v_2)\cdot A, ~(w_1 + w_2 + v_1 \wedge v_2)\cdot\Lambda^2(A) + (v_1+v_2)\cdot b ) \\ = (v_1+v_2, ~w_1+w_2+v_1\wedge v_2)M \\ = ( (v_1,w_1)\cdot(v_2,w_2) )M \end{array}$$

The composition of homomorphisms M and M′ is given by matrix multiplication: $$M\cdot M' = \begin{bmatrix} A & b \\ . & \Lambda^2(A) \end{bmatrix} \cdot \begin{bmatrix} A' & b' \\ . & \Lambda^2(A') \end{bmatrix} = \begin{bmatrix} A\cdot A' & A\cdot b' + b \cdot \Lambda^2(A') \\ . & \Lambda^2(A\cdot A')\end{bmatrix}$$

When dim(V)=2, it would be nice to use parameters (A,u) to describe M, but the question is how to multiply in these parameters (A,u)⋅(A′,u′) is not simply (A⋅A′, Au′+u⋅det(A′)). Here A∧u′ is the element of Hom(V,W) which takes v in V to (vA)∧u′ in W. We can in fact simplify A∧u′ to just ∧Bu′ for some matrix B related to A.

Unfortunately, A ≠ B in general. Instead, one gets B = A−T ⋅ det(A), I believe. In other words, up to a determinant, we get the action of A on its dual module rather than on its natural module. I'd like a clearer description of this action, since I am pretty certain the real answer has the natural module. Hopefully, there is another dual I am forgetting that makes everything ok. At any rate, I have no trouble doing it with coordinates as below:

It is comforting to put coordinates on the group: [x,y,z] = ( x⋅e1 + y⋅e2, z⋅e1∧e2 ). As matrices, this can be realized as: $$[x,y,z] = \begin{pmatrix} 1 & x & z + xy \\ . & 1 & 2y \\ . & . & 1 \end{pmatrix}$$

Multiplication follows the rule [a,b,c]⋅[x,y,z] = [a+x,b+y,c+z+ay-bx].

Powers obey the very simple rule [x,y,z]n = [ nx, ny, nz ].

Commutators obey the simple rule [ [a,b,c], [x,y,z] ] = [ 0, 0, 2(ay-bx) ].

Now consider a matrix: $$A = \begin{pmatrix} a & b & e \\ c & d & f \\ . & . & ad-bc \end{pmatrix}$$

One can check that in these coordinates A is an automorphism. In particular, the function that takes the row vector [x,y,z] times the matrix A to get a new row vector [x′, y′, z′ ] defines an automorphism of G. Composition of automorphisms is matrix multiplication. In this coordinate system everything is perfect.

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Dear Jack, I would like to ask a small question: can u explain, how the maps which takes x to x^a.y^b, y to x^c.y^d (ad-bc=/=0), form subgroup GL(2,p)? Means my question (may be easy, but i too puzzled) is , does the composition of two such maps gives same type of map? – user8186 Jan 28 '11 at 4:13
I think I've left out a factor of 2 somewhere. If you choose "z" right for the automorphisms, then composition of automorphisms is matrix multiplication, and application to group elements is a vec-mat multiplication. The version right now gets cluttered with binomial coefficients and determinants, but I remember there is a natural basis where everything works out reasonably cleanly. – Jack Schmidt Jan 28 '11 at 5:34
Rahul and Alex: I've removed both of your comments for being off topic. – Willie Wong Jan 28 '11 at 19:05
Just want to mention I wrote up a PDF describing what is going on here for Rahul's subsequent question on MO. It is essentially what Jack is talking about, though I give an explicit realization of GL(2,p) in the automorphism group. It can be found at math.sunysb.edu/~sdalton/aut_p3_groups.pdf [Jack's comment above is to a comment I posted and subsequently merged into this one.] – user641 Mar 13 '11 at 21:18
@Steve: Cool, I put up my current version using the wedgey group. The coordinate-free approach is still buggy, but with coordinates everything is fine. You might find the mysterious factors of 2 I mentioned (and you included as 1/2 in the automorphisms) more natural in wedgey land. I don't like the matrix realization of the p-group though; it just looks silly. – Jack Schmidt Mar 14 '11 at 0:01