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 .

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## Comments

buenas noches quisiera felicitarlos por la pagina y ademas les recomiendo que vean este video en el cual se demuestra el ejercicio 7 de la secion 3.3 es otra forma

Thank you for reading, and thank you for the link to this video! His proof is much more straightforward than mine- I have a tendency to immediately reach for the First Isomorphism Theorem when defining homomorphisms from a quotient.

Hi,

I think we still can make use of 1st Homom Thm and define this map:

\Phi: AxB -> (A/C) x (B/D)

by

\Phi((a,b)) = (aC,bD).

And, show that \Phi is a surjective homom and Ker \Phi = C x D.

I guess this proof is much simpler.

This works and is true, but the problem here is a tad more general- is not necessarily the

directproduct of and .in your proof did you forget to prove its a homomorphism?

You’re right! I’ll fix it.

How about this? It seems more simple!

Let by . It is easy to see that is a homomorphism and . To show the surjectivity, let be any element of . Then . If we choose , , as desired.

Wow- that’s much better!

I’ll leave the old proof up for comparison. Thanks!