Definition. Let R be a commutative ring with 1. The Heisenberg group H(R) is defined by

H(R) = \left \{ \begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}: \ \ a,b,c \in R \right \}.

Clearly H(R) is a group under matrix multiplication. Thus H(R) is a subgroup of the group of 3 \times 3 invertible upper triangular matrices with entries from R.

Problem 1. Let p be a prime number and let R=\mathbb{F}_p, the field of order p. Let G be the group of 3 \times 3 invertible upper triangular matrices with entries from R. Prove that H(R) is the unique Sylow p-subgroup of G.

Solution. Let g = \begin{pmatrix} x & y & z \\ 0 & u & v \\ 0 & 0 & w \end{pmatrix} \in G. Since g is invertible we must have xuw \neq 0 and so x, u, w \neq 0. Thus, since the entries of G come from R, there are p-1 possible values for x, u, w and p possible values for y,z,v. Thus |G|=p^3(p-1)^3. It is obvious that |H(R)|=p^3. Since \gcd(p,p-1)=1, we see that H(R) is indeed a Sylow p-subgroup of G. To show that H(R) is the only Sylow p-subgroup of G, we only need to show that H(R) is normal in G because, by Sylow theorem, Sylow p-subgroups are conjugates. To prove that H(R) is normal in G, suppose that R^{\times} is the multiplicative group of R. Let K = R^{\times} \times R^{\times} \times R^{\times}. Define G \longrightarrow K by

\varphi(g) = (x,u,w),

for all g =\begin{pmatrix} x & y & z \\ 0 & u & v \\ 0 & 0 & w \end{pmatrix} \in G. It is straightforward to check that \varphi is a group homomorphism and \ker \varphi = H(R). Thus H(R) is a normal subgroup of G. \ \Box

Notation. Let R be a commutative ring with 1. For the sake of simplicity, let’s use this notation

[a,b,c] = \begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}.

So H(R)=\{[a,c,c] : \ a,b,c \in R \}.

Remark.  Multiplying two elements of H(R) will gives us the following identities

[a,b,c][a',b',c']=[a+a', b+b', c+c'+ab'], \ \ [a,b,c]^{-1} = [-a,-b,ab-c].

Problem 2. Let R be a commutative ring with 1 and suppose that A = \{[0, b, c] : \ \ b,c \in R \} and B = \{[a,0,0]: \ \ a \in R \}. Prove that

1) A and B are subgroups of H(R) and A is normal in H(R).

2) H(R)=A \rtimes B, i.e. H(R) is the semidirect product of A and B.

Solution. 1) Let x=[0,b,c], \ y = [0,b',c'] be two elements of A. Then by the above remark

xy^{-1}=[0,b,c][0,-b',-c']=[0, b-b',c-c'] \in A.

So A is a subgroup of H(R). Similarly, B is a subgroup of H(R). Now, let g = [r,s,t] be any element of H(R). Then gxg^{-1} = [r,s,t][0,b,c][-r,-s,rs-t]=[0,b,c+rb] \in A. So A is normal in H(R). Note that gyg^{-1} \notin B in general and so B is not normal in H(R).

2) As we showed in the first part, A is normal in H(R) and clearly A \cap B = \{1\} and H(R)=AB. \ \Box

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