
What is a Commutator? - BYJU'S
What is a Commutator? Commutators are used in DC machines (DC motors and DC generators) universal motors. In a motor, a commutator applies an electric current to the windings. A …
What is a commutator - Mathematics Stack Exchange
The second way is to look at the commutator subgroup as a measure of how noncommutative a group is. A group is commutative if it has a trivial commutator subgroup (and highly …
How to show that the commutator subgroup is a normal subgroup
The commutator subgroup is generated by commutators. Show that the property of "being a commutator" is invariant under conjuation (in fact it is invariant under all automorphisms).
Understanding the commutator subgroup of the dihedral group
@NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
The commutator of two matrices - Mathematics Stack Exchange
The commutator [X, Y] of two matrices is defined by the equation $$\begin {align} [X, Y] = XY − YX. \end {align}$$ Two anti-commuting matrices A and B satisfy $$\begin {align} A^2=I \qu...
Center-commutator duality - Mathematics Stack Exchange
So here's a sense in which the commutator subgroup and the center are "dual": the commutator is the subgroup generated by all values of $\mathbf {w} (x,y)$, and the center is the subgroup of …
Why is the commutator defined differently for groups and rings?
Jun 30, 2015 · The commutator of a group and a commutator of a ring, though similar, are fundamentally different, as you say. In each case, however, the commutator measures the …
Commutator Group of $\\operatorname{GL}_2(\\mathbb{R})$ is ...
May 2, 2018 · Let $\operatorname {GL}_2 (\mathbb {R})$ be the general linear group of $2\times2$ matrices and $\operatorname {SL}_2 (\mathbb {R})$ the special linear group of $2 …
Commutator of $p^2$ and $x$ - Mathematics Stack Exchange
Jan 23, 2020 · Writing in more details, the commutator relation is $$ [p,x]=-i\hbar I$$ where $I$ is the identity operator. So $pi\hbar$ should be understood as $pi\hbar I$, the composition of the …
Calculating the commutator (derived) subgroup of $S_3$
If $x$ and $y$ are in $S_3$, then their commutator, $x^ {-1}y^ {-1}xy$, is an even permutation. So the commutator subgroup is a subgroup of $A_3$, which is just the identity and the 3-cycles.