A two-digit link on a set fulfills the Moufang identities (named after the German mathematician Ruth Moufang ), if the equations
for all${\ displaystyle \ cdot}$${\ displaystyle X}$${\ displaystyle a, b, c \ in X}$

(M1) ${\ displaystyle {\ Big (} a \ cdot (b \ cdot a) {\ Big)} \ cdot c = a \ cdot {\ Big (} b \ cdot (a \ cdot c) {\ Big)}}$

and

(M2) ${\ displaystyle (a \ cdot b) \ cdot (c \ cdot a) = a \ cdot {\ Big (} (b \ cdot c) \ cdot a {\ Big)}}$

be valid.

The following equations are also referred to as Moufang identities:

(M1 ') ${\ displaystyle {\ Big (} (a \ cdot b) \ cdot c {\ Big)} \ cdot b = a \ cdot {\ Big (} b \ cdot (c \ cdot b) {\ Big)}}$

and

(M2 ') ${\ displaystyle (a \ cdot b) \ cdot (c \ cdot a) = {\ Big (} a \ cdot (b \ cdot c) {\ Big)} \ cdot a}$

In a quasi-group , one of these four equations implies the other three. In addition, each of these equations assures the existence of a neutral element . A quasi-group, in which (at least) one of the Moufang identities is fulfilled, is therefore a loop , which is then also called a Moufang loop .
${\ displaystyle (M, \ cdot)}$

If the Moufang identities (M1) and (M2) apply in a magma with a neutral element , then the link applies${\ displaystyle (M, \ cdot)}$${\ displaystyle 1}$${\ displaystyle \ cdot}$

the left alternative (because of (M1) with ):${\ displaystyle b = 1}$

${\ displaystyle (a \ cdot a) \ cdot c = {\ Big (} a \ cdot (1 \ cdot a) {\ Big)} \ cdot c = a \ cdot {\ Big (} 1 \ cdot (a \ cdot c) {\ Big)} = a \ cdot (a \ cdot c)}$

the flexibility law (because of (M2) with ):${\ displaystyle c = 1}$

${\ displaystyle (a \ cdot b) \ cdot a = (a \ cdot b) \ cdot (1 \ cdot a) = a \ cdot {\ Big (} (b \ cdot 1) \ cdot a {\ Big)} = a \ cdot (b \ cdot a)}$

Apply in a neutral element , however, 'and (M2 Moufang identities (M1)'), then for the combination${\ displaystyle (M, \ cdot)}$${\ displaystyle 1}$${\ displaystyle \ cdot}$

${\ displaystyle (a \ cdot b) \ cdot b = {\ Big (} (a \ cdot b) \ cdot 1 {\ Big)} \ cdot b = a \ cdot {\ Big (} b \ cdot (1 \ cdot b) {\ Big)} = a \ cdot (b \ cdot b)}$

the flexibility law (because of (M2 ') with ):${\ displaystyle b = 1}$

${\ displaystyle a \ cdot (a \ cdot b) = (a \ cdot 1) \ cdot (c \ cdot a) = {\ Big (} a \ cdot (1 \ cdot c) {\ Big)} \ cdot a = (a \ cdot b) \ cdot a}$

In a flexible magma , in which the law of flexibility applies to the link , M2 'follows directly from M2 (and vice versa), and the following additional identities apply
${\ displaystyle (M, \ cdot)}$${\ displaystyle \ cdot}$

(M3, follows from M1) ${\ displaystyle {\ Big (} (a \ cdot b) \ cdot a {\ Big)} \ cdot c = {\ Big (} a \ cdot (b \ cdot a) {\ Big)} \ cdot c = a \ cdot {\ Big (} b \ cdot (a \ cdot c) {\ Big)}}$

(M3 ', follows from M1') ${\ displaystyle {\ Big (} (a \ cdot b) \ cdot c {\ Big)} \ cdot b = a \ cdot {\ Big (} b \ cdot (c \ cdot b) {\ Big)} = a \ cdot {\ Big (} (b \ cdot c) \ cdot b {\ Big)}}$