This topic is linked to null - null.
Y by ishan.panpaliya, felixgotti
What I've tried so far:
(1) If are two distinct identity elements of , then , a contradiction.
(2) Let be the identity element of . If are two distinct inverses of , then
a contradiction.
(3) Claim 1: Associativity holds in
.
Claim 2: Identity holds in
The identity element of is the inverse of itself, so it is also in . Since the operations in and are the same, it is the identity of as well.
Claim 3: Closure holds in
Let the identity element be .
.
, so is invertible (namely, its inverse is ), which implies .
Claim 4: Every element in has an inverse
This is true by the definition of .
By definition, these four claims are sufficient to determine is a group.
Where I'm stuck:
Please check
(1) If are two distinct identity elements of , then , a contradiction.
(2) Let be the identity element of . If are two distinct inverses of , then
a contradiction.
(3) Claim 1: Associativity holds in
.
Claim 2: Identity holds in
The identity element of is the inverse of itself, so it is also in . Since the operations in and are the same, it is the identity of as well.
Claim 3: Closure holds in
Let the identity element be .
.
, so is invertible (namely, its inverse is ), which implies .
Claim 4: Every element in has an inverse
This is true by the definition of .
By definition, these four claims are sufficient to determine is a group.
Where I'm stuck:
Please check