Kernel (algebra)
Proof: If f were not injective, then the non-injective elements can form a distinct element of its kernel: there would exist  such that  and . Thus . f is a group homomorphism, so inverses and group operations are preserved, giving ; in other words, , and ker f would not be the singleton. Conversely, distinct elements of the kernel violate injectivity directly: if there would exist an element , then , thus f would not be injective.