Let be a group and let be normal such that . Prove that .
[New, short proof]
Define by ; certainly is a homomorphism. If , then , so that . Conversely, is contained in .
Now we argue that is surjective. To see this, let . Since , and since and are normal, we have for some and . Let ; now for some , and evidently . So is surjective. By the first isomorphism theorem, we have .
[Old, long proof]
Note that every has the form for some and . Define a mapping as follows. If , then .
Well defined: Suppose . Now , so that . Thus , and we have . Similarly, . Thus is well defined.
Homomorphism: Let . Note that, since is normal in , we have . Note also that . Since and are both normal in , this product is in . So we have . Now . So is a group homomorphism.
Surjective: If , we have . Thus is surjective.
: Suppose . Then , so that . Hence . Suppose . Now , so that . Thus .
By the First Isomorphism Theorem, we have . Moreover by the Second isomorphism theorem we have and . Thus .