Group algebra This is the same idea as matrix algebra in FP1.Ex1 a • b = c make b the subject a -1• on left: a -1 • (a • b) = a-1 • c use associative rule (a -1• a) • b = a-1 • c use inverse rule e • b = a-1 • c use identity rule b = a-1 • c

Ex2 Prove (a • b • c)-1 = c-1 • b-1 • a-1 Use inverse rule (a • b • c)-1 • (a • b • c) = e •c -1 on right: (a • b • c)-1 • a • b • c • c -1 = e • c -1 inverse and identity rules: (a • b • c)-1 • a • b • e = e • c -1 (a • b • c)-1 • a • b = c -1 •b -1 on right: (a • b • c)-1 • a • b • b-1= c -1 • b-1 inverse and identity rules: (a • b • c)-1 • a • e = c -1 • b-1 (a • b • c)-1 • a = c -1 • b-1

•a -1 on right: (a • b • c)-1 • a • a -1 = c -1 • b-1 • a -1 inverse and identity rules: (a • b • c)-1 • e = c -1 • b-1 • a -1 (a • b • c)-1 = c -1 • b-1 • a -1

Ex3Prove that if c commutes with every element of a group then c-1 does too. Let x be any element in the group. We are told c • x = x • c and the objective is to prove that c-1• x = x • c-1

c • x = x • c

•c -1 on right left and c -1• on right : c-1• c • x • c-1 = c-1 • x • c • c-1

e • x • c -1 = c -1 • x • e x • c -1 = c -1 • x as required

Ex4 Find an inverse of a in R–{1} with the binary operation a • b = a + b – abR–{1} means the set of real numbers excluding 1

1) Find the identity elementa • e = a hence using a • b = a + b – ab

a • e = a + e – ae replace b by ea • e = a + e – a as ae = aa • e = e

ae – e = 0 e(a – 1) = 0 as a is an element in R–{1} a 1 So e = 0

Self-inversive groups A system of algebra exists for every group, which can be used to make surprising conclusions:e.g. If a group X is self - inversive then x2 = e x X Now look at x • y: x • y = e •( x • y )• e as this changes nothing by identity = y • y • x • y • x • x as x • x = y • y = e for this group = y • (y • x) • (y • x) • x by associativity = y • e • x as (y • x) X by closure = y • xConclusion: Self-inversive groups are commutative

Using contradictionEx1 If x , y and the identity element e are elements of a group H and x y = y2 x prove that x y y x Proof Assume x y = y xso y x = y2 x = y y xPost multiply by (y x)–1 (y x) (y x)–1 = y (y x) (y x)–1

e = y e = ybut y e so x y y x