## Heisenberg group

Posted: February 14, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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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 $H(R)=AB. \ \Box$