Let \(G\) be the group of matrices of the form \[\begin{align*} \begin{pmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{pmatrix} .\end{align*}\]
with entries in the finite field \(\mathbb F_p\) of \(p\) element, where \(p\) is a prime.
Prove that \(G\) is non-abelian.
Suppose \(p\) is odd. Prove that \(g^p=I_3\) for all \(g\in G\).
Suppose that \(p=2\). It is known that there are exactly two non-abelian groups of order 8, up to isomorphism: the dihedral group \(D_8\) and the quaternionic group. Assuming this fact without proof, determine which of these groups \(G\) is isomorphic to.
Solution
1. \(G\) is the group of upper-triangular unipotent \(3 \times 3\) matrices over \(\mathbb{F}_p\), the Heisenberg group over \(\mathbb{F}_p\).
the matrices \(\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}\) form the Heisenberg group.
2. \(G\) is non-abelian.
e.g. \(\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\) do not commute (their commutator is \(\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \neq I\)).
3. Write \(g = I + N\) where \(N = \begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix}\) is strictly upper-triangular, so \(N^3 = 0\).
strictly upper-triangular \(3 \times 3\) matrices are nilpotent of index \(3\).
4. \(g^p = (I + N)^p = I + pN + \binom{p}{2}N^2 + \cdots\).
binomial theorem, and \(N^3 = 0\) kills all higher terms.
5. If \(p\) is odd, then \(p \equiv 0 \pmod p\) and \(\binom{p}{2} = \frac{p(p-1)}{2} \equiv 0 \pmod p\) (since \(p\) is odd, \(p-1\) is even, so \(\binom{p}{2}\) is divisible by \(p\)).
arithmetic in \(\mathbb{F}_p\).
6. Hence \(g^p = I\) for all \(g \in G\) when \(p\) is odd.
4 and 5 (all binomial coefficients \(\binom{p}{k}\) for \(1 \le k \le p-1\) are divisible by \(p\)).
7. For \(p = 2\), \(G\) has order \(2^3 = 8\).
there are \(p^3 = 8\) choices of \((a, b, c) \in \mathbb{F}_2^3\).
8. For \(p = 2\), \(G\) is non-abelian of order \(8\), so it is either \(D_8\) or the quaternion group \(Q_8\).
2, 7, and the given fact.
9. \(G\) has more than one element of order \(2\) (e.g. \(\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\) both have order \(2\)), whereas \(Q_8\) has a unique element of order \(2\).
in \(Q_8\), only \(-1\) has order \(2\); in \(G\), the two matrices above (and more) have order \(2\).
10. Hence \(G \cong D_8\) (the dihedral group of order \(8\)).
8 and 9.
11. Q.E.D.
2, 6, 10.