Commutative Subsets of an Algebra

# Commutative Subsets of an Algebra

Definition: Let $X$ be an algebra. A subset $E \subseteq X$ is said to be a Commutative Subset of $X$ if $xy = yx$ for all $x, y \in E$. |

Definition: Let $X$ be an algebra and let $E \subseteq X$. The Commutant of $E$ is the set $E^c = \{ x \in X : ex = xe \: \forall e \in E \}$. The Second Commutant of $E$ is the set $E^{cc} := (E^c)^c = \{ x \in X : e'x = xe' \: \forall e' \in E \}$. The Centre of $X$ is the commutant of $X$, i.e., $X^c$. |

Proposition 1: Let $X$ be an algebra and let $E \subseteq X$. Then $E^c$ is a subalgebra of $X$. Moreover, if $X$ is an algebra with unit $1$ then $E^c$ is a subalgebra with unit $1$. _ |

**Proof:**Let $x, y \in E^c$. Then $ex = xe$ and $ey = ye$ for all $e \in E$. So $(x + y)e = xe + ye = ex + ey = e(x + y)$ for all $e \in E$, so $x + y \in E^c$ whenever $x, y \in E^c$, showing that $E^c$ is closed under addition.

- Let $\alpha \in \mathbf{F}$ and let $x \in E^c$. Then $ex = xe$ for all $e \in E$. So $e(\alpha x) = \alpha(ex) = \alpha(xe) = (\alpha x)e$ for all $e \in E$. Therefore $\alpha x \in E^c$ whenever $\alpha \in \mathbf{F}$ and $x \in E^c$, showing that $E^c$ is closed under scalar multiplication.

- Let $x, y \in E^c$. Then $ex = xe$ and $ey = ye$ for all $e \in E$ again. So $e(xy) = (ex)y = (xe)y = x(ey) = x(ye) = (xy)e$ for all $e \in E$. Therefore $xy \in E^c$ whenever $x, y \in E^c$, showing that $E^c$ is closed under multiplication.

- Finally, $0e = 0 = e0$ for all $e \in E$ and so $0 \in E^c$. So $E^c$ is a subalgebra of $X$.

- If $X$ is an algebra with unit $1$ then by definition, $x1 = x = 1x$ for all $x \in X$. In particular, $e1 = e = 1e$ for all $e \in E$ and so $1 \in E^c$. So if $X$ is an algebra with unit $1$ then $E^c$ is a subalgebra with unit $1$. $\blacksquare$

Proposition 2: Let $X$ be an algebra and let $E \subseteq X$. Then $E$ is a commutative subset of $X$ if and only if $E \subseteq E^c$. |

**Proof:**$\Rightarrow$ Suppose that $E$ is a commutative subset of $X$. Let $x \in E$. Then $xy = yx$ for all $y \in E$. So $x \in E^c$ showing that $E \subseteq E^c$.

- $\Leftarrow$ Conversely, suppose that $E \subseteq E^c$. Let $x, y \in E$. Then $x, y \in E^c$. Since $x \in E^c$ and $y \in E$ we have that $xy = yx$. Since this holds true for all $x, y \in E$ we see that $E$ is a commutative subset of $X$.

Proposition 3: Let $X$ be an algebra and let $E, F \subseteq X$. If $E \subseteq F$ then $F^c \subseteq E^c$. |

**Proof:**Suppose that $E \subseteq F$. Let $x \in F^c$. Then $xf = fx$ for all $f \in E$. Since $E \subseteq F$ this tells us that $xe = ex$ for all $e \in E$. So $x \in E^c$. Thus $F^c \subseteq E^c$. $\blacksquare$

Proposition 4: Let $X$ be an algebra and let $E \subseteq X$. If $E$ is a commutative subset of $X$ then $E^{cc}$ is a commutative subalgebra of $X$ and $E \subseteq E^{cc} \subseteq E^c$; |

**Proof:**By proposition 1, $E^c$ is a subalgebra of $X$. Applying this proposition again and we see that $E^{cc}$ is a subalgebra of $E^c$, and so $E^{cc}$ is a subalgebra of $X$.

- Now since $E$ is a commutative subset of $X$ we have by proposition 2 that $E \subseteq E^c$. So by proposition 3 we have that $E^{cc} \subseteq E^c$.

- Now let $x \in E$. Let $y \in E^c$. Then $ey = ye$ for all $e \in E$. In particular, $xy = yx$ for all $y \in E^c$. Thus $x \in E^{cc}$ and hence $E \subseteq E^{cc}$. Combining the above two set inclusions gives us that $E \subseteq E^{cc} \subseteq E$.

- Lastly, let $x, y \in E^{cc}$. Then from above since $E^{cc} \subseteq E^c$ we have that $x, y \in E^c$. So for $x \in E^{cc}$ and $y \in E^c$ we have by definition of $x \in E^{cc}$ that $xy = yx$. So $E^{cc}$ is a commutative subalgebra of $X$. $\blacksquare$

Proposition 5: Let $X$ be an algebra. Then the centre of $X$ is a commutative subalgebra of $X$. |

**Proof:**By proposition 1, $X^c$ is a subalgebra of $X$. Let $x, y \in X^c$. Since $x \in X^c$ by definition we have that $xz = zx$ for all $z \in X$. In particular, $xy = yx$ since $y \in X$. So $xy = yx$ for all $x, y \in X^c$ and thus $X^c$ is a commutative subset of $X$. By proposition 4, $X^c$ is a commutative subalgebra of $X$. $\blacksquare$