Let be a group homomorphism. Define the kernel of to be . Prove that is a subgroup of . Prove that is injective if and only if .
By a previous example, it suffices to show that is closed under multiplication and inversion. To that end, let . Then , so that . Now let . Then so that . Thus is a subgroup of .
Now we prove the second statement.
Suppose is injective, and let . Note that , so that . Thus .
Suppose , and let such that . Now , so that and thus . Hence is injective.