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The Heisenberg multiplication is a group law, nonabelian, on a set of $p^{3}$ elements

Statement

Let $p$ be a prime. The multiplication $(a, b, c) (a^{'}, b^{'}, c^{'}) = (a + a^{'}, b + b^{'}, c + c^{'} + a b^{'})$ makes the set $H_{p}$ a group with identity $(0, 0, 0)$ and inverse $(a, b, c)^{- 1} = (- a, - b, - c + a b)$ ; the group is not abelian; and $∣ H_{p} ∣ = p^{3}$ . Moreover $(1, 0, 0)^{a} = (a, 0, 0)$ , $(0, 1, 0)^{b} = (0, b, 0)$ and $(0, 0, 1)^{c} = (0, 0, c)$ for all $a, b, c$ , so those three elements generate $H_{p}$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups) and each has order $p$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

Facts & Assumptions

Given: A prime $p$ and the set $H_{p}$ with the multiplication above.

[F1]

The Heisenberg group of order $p^{3}$ is the set $H_{p} = {(a, b, c) : a, b, c \in Z / p}$ with $(a, b, c) (a^{'}, b^{'}, c^{'}) : = (a + a^{'}, b + b^{'}, c + c^{'} + a b^{'})$ (The Heisenberg group of order $p^{3}$ over $Z / p$ ).

[F2]

A group is a set with an associative operation having a two-sided identity and two-sided inverses (Group and abelian group).

[L1]

For every $n \in N$ , $(Z / n, +, [0]_{n})$ is an abelian group, $(Z / n, \cdot, [1]_{n})$ is a commutative monoid, and multiplication distributes over addition on both sides (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L2]

If $A$ and $B$ are finite then $A \times B$ is finite and $∣ A \times B ∣ = ∣ A ∣ \cdot ∣ B ∣$ (The product rule: $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ , and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ ).

[L3]

$∣ A ∣$ is the unique natural number $n$ with $A \approx n$ (The cardinality $∣ A ∣$ of a finite set).

[L4]

For every prime $p$ , $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

Proof

technique · direct

1.1

Associativity: both $((a, b, c) (a^{'}, b^{'}, c^{'})) (a^{''}, b^{''}, c^{''})$ and $(a, b, c) ((a^{'}, b^{'}, c^{'}) (a^{''}, b^{''}, c^{''}))$ have first coordinate $a + a^{'} + a^{''}$ , second coordinate $b + b^{'} + b^{''}$ , and third coordinate $c + c^{'} + c^{''} + a b^{'} + a b^{''} + a^{'} b^{''}$ , the two computations differing only in the order in which the three products are formed.

F1L1algebra

1.2

The triple $(0, 0, 0)$ is a two-sided identity: $(a, b, c) (0, 0, 0) = (a, b, c + 0 + a \cdot 0) = (a, b, c)$ and $(0, 0, 0) (a, b, c) = (a, b, 0 + c + 0 \cdot b) = (a, b, c)$ .

F1L1algebra

1.3

The triple $(- a, - b, - c + a b)$ is a two-sided inverse of $(a, b, c)$ : the product in one order is $(0, 0, c + (- c + a b) + a (- b)) = (0, 0, 0)$ , and in the other it is $(0, 0, (- c + a b) + c + (- a) b) = (0, 0, 0)$ .

F1L1algebra

1.4

The quotient set $Z / p$ has the $p$ distinct classes $[0]_{p}, \dots, [p - 1]_{p}$ , so it has $p$ elements. The underlying set of $H_{p}$ is the threefold product of those $p$ -element sets, and [L2] therefore gives $∣ H_{p} ∣ = p^{3}$ .

F1L2L3algebra

1.5

The three displayed power formulas hold because $(a, 0, 0) (1, 0, 0) = (a + 1, 0, 0 + 0 + a \cdot 0) = (a + 1, 0, 0)$ , $(0, b, 0) (0, 1, 0) = (0, b + 1, 0 + 0 + 0 \cdot 1) = (0, b + 1, 0)$ and $(0, 0, c) (0, 0, 1) = (0, 0, c + 1)$ , so each power is obtained from the previous one by adding one in the relevant coordinate.

F1L1algebra

2.1

By steps 1.1 to 1.3 the multiplication makes $H_{p}$ a group.

F2step 1.1step 1.2step 1.3

2.2

It is not abelian: $(1, 0, 0) (0, 1, 0) = (1, 1, 1)$ while $(0, 1, 0) (1, 0, 0) = (1, 1, 0)$ , and these differ because $1 \neq 0$ in $Z / p$ for every prime $p$ .

F1L4step 1.4

3.1

Each of $(1, 0, 0)$ , $(0, 1, 0)$ and $(0, 0, 1)$ has $p$ -th power $(0, 0, 0)$ by step 1.5, hence order $p$ since each is not the identity; and $(a, 0, 0) (0, b, 0) (0, 0, c - a b) = (a, b, a b) (0, 0, c - a b) = (a, b, c)$ , so the three elements generate $H_{p}$ .

L1step 1.5step 2.1step 2.2∎

Remarks

The verification of associativity is where the third coordinate earns its shape: the two bracketings produce the cross terms $a b^{'} + a b^{''}$ and $a b^{'} + a b^{''}$ respectively together with the common term $a^{'} b^{''}$ , and they agree because multiplication in $Z / p$ distributes over addition.

Depends on

Used by

Cited to discharge well-definedness by The Heisenberg group of order p³ over ℤ/p.

Dependency tree · two levels

45 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

M. van Beek, Topics in Finite p-Groups, Definition 2.31 (standard reference, not scraped)
D. A. Craven, The Theory of p-Groups, Definition 3.3 (standard reference, not scraped)

The Heisenberg multiplication is a group law, nonabelian, on a set of p3p3 elements

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Used by

Dependency tree · two levels

Sources

The Heisenberg multiplication is a group law, nonabelian, on a set of $p^{3}$ elements