# Frobenius's Theorem

## Theorem

An algebraic associative real division algebra $A$ is isomorphic to $\R, \C$ or $\Bbb H$.

## Proof

Recall that an algebra $A$ is said to be quadratic if it is unital and the set $\left\{{1, x, x^2}\right\}$ is linearly dependent for every $x \in A$.

Thus, for every $x \in A$ there exist $t \left({x}\right), n \left({x}\right) \in \R$ such that:

- $x^2 - t \left({x}\right) x + n \left({x}\right) = 0$

Obviously, $t \left({x}\right)$ and $n \left({x}\right)$ are uniquely determined if $x \notin \R$.

Suppose $x \notin \R$.

Then $x$ can be expressed as $x = a + b i$, with $a, b \in \R$ and $b \ne 0$.

Then:

- $x^2 = a^2 - b^2 + 2 a b i$

and:

- $x^2 - t \left({x}\right) x + n \left({x}\right) = a^2 - b^2 - a t \left({x}\right) + n \left({x}\right) + \left({2 a b - b t \left({x}\right)}\right) i = 0$

It follows that both:

- $(1): \quad 2 a b - b t \left({x}\right) = 0$

and:

- $(2): \quad a^2 - b^2 - a t \left({x}\right) + n \left({x}\right) = 0$

$(1)$ leads to:

- $t \left({x}\right) = 2 a$

and $(2)$ leads to;

- $n \left({x}\right) = a^2 + b^2$

Setting $t \left({\lambda}\right) = 2 \lambda$ and $n \left({\lambda}\right) = \lambda^2$ for $\lambda \in \R$, we can then consider $t$ and $n$ as maps from $A$ into $\R$.

(In this way $t$ becomes a linear functional).

We call $t \left({x}\right)$ and $n \left({x}\right)$ the trace and the norm of $x$ respectively.

From $x^2 - \left({x + x^*}\right) x + x^* x = 0$ we see that all algebras $\Bbb A_n$ are quadratic.

Further, every real division algebra $A$ that is algebraic and power-associative (this means that every subalgebra generated by one element is associative) is automatically quadratic.

Indeed, if $x \in A$ then there exists a nonzero polynomial $f \left({X}\right) \in \R \left[{X}\right]$ such that $f \left({x}\right) = 0$.

Writing $f \left({X}\right)$ as the product of linear and quadratic polynomials in $\R \left[{X}\right]$ it follows that $p \left({x}\right) = 0$ for some $p \left({X}\right) \in \R \left[{X}\right]$ of degree $1$ or $2$.

In particular, algebraic alternative (and hence associative) real division algebras are quadratic.

Finally, if $A$ is a real unital algebra, that is, an algebra over $\R$ with unity $1$, then we shall follow a standard convention and identify $\R$ with $\R 1$.

Thus we shall write $\lambda$ for $\lambda 1$, where $\lambda \in \R$.

### Lemma 1

Let $\left({A, \oplus}\right)$ be a quadratic real algebra.

Then:

- $(1): \quad U = \left\{{u \in A \setminus \R: u^2 \in \R}\right\} \cup \left\{{0}\right\}$ is a linear subspace of $A$

- $(2): \quad \forall u, v \in U: u v + v u \in \R$

- $(3): \quad A = \R \oplus U$

- $(4): \quad$ If $A$ is also a division algebra, then every nonzero $u \in U$ can be written as $u = \alpha v$ with $\alpha \in \R$ and $v^2 = -1$.

### Lemma 2

Let $A$ be a quadratic real division algebra.

Let:

- $U = \left\{{u \in A \setminus \R: u^2 \in \R}\right\} \cup \left\{{0}\right\}$

where $\setminus$ denotes set difference.

Suppose $e_1, \ldots, e_k \in U$ are such that:

- $\forall i \le k: e_i^2 = -1$
- $\forall i, j \le k, i \ne j: e_i e_j = -e_j e_i$

If $U$ is not equal to the linear span of $e_1, \ldots, e_k$, then there exists $e_{k+1} \in U$ such that:

- $e_{k+1}^2 = -1$

- $\forall i \le k: e_i e_{k+1} = -e_{k+1} e_i$

We have from above that $A$ is quadratic.

We may assume that $n = \dim A \ge 2$.

By Lemma 1: Assertion 4 we can fix $i \in A$ such that $i^2 = -1$.

Thus, $A \cong \C$ if $n = 2$.

Let $n > 2$.

By Lemma 2:

- $\exists j \in A: j^2 = -1, i j = -j i$

Set $k = ij$.

It can immediately be checked that:

- $k^2 = -1$
- $k i = j = -i k$
- $j k = i = -k j$
- $\left\{{i, j, k}\right\}$ is a linearly independent set.

Therefore $A$ contains a subalgebra isomorphic to $\Bbb H$.

Finally, suppose $n > 4$.

By Lemma 2 there would exist $e \in A, e \ne 0$ such that:

- $(a): \quad e i = -i e$
- $(b): \quad e j = -j e$
- $(c): \quad e k = -k e$

However, from the $(a)$ and $(b)$ it follows that $e i j = -i e j = i j e$.

Since $i j = k$, this contradicts $(c)$.

It follows that $n \le 4$, and so $\Bbb H$ is the highest order

$\blacksquare$

## Source of Name

This entry was named for Ferdinand Georg Frobenius.

## Historical Note

Frobenius's Theorem was proved by Ferdinand Georg Frobenius in $1878$.

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem - Matej Brešar, Peter Šemrl and Špela Špenko:
*On Locally Complex Algebras and Low-Dimensional Cayley-Dickson Algebras*(