1. ## proving surjective homomorphism

Requesting a bit of help on the last part of my homework.

Need to prove that the following is a surjective homomorphism:

Let H and K be groups. Define the projection map:

: H x K H such that : (h, k) h

: H x K K such that : (h, k) k

How do you prove something is onto?

I know for a homomorphism you show that:
(ab) = (a)(b)

How would I attack this proof?

Thanks.

2. Originally Posted by kpizle
Requesting a bit of help on the last part of my homework.

Need to prove that the following is a surjective homomorphism:

Let H and K be groups. Define the projection map:

: H x K H such that : (h, k) h

: H x K K such that : (h, k) k

How do you prove something is onto?

I know for a homomorphism you show that:
(ab) = (a)(b)

How would I attack this proof?

Thanks.

I'm guessing you must prove $\displaystyle \pi_1\,,\,\pi_2$ are surjective. Take the first one: since $\displaystyle H\times K=\{(h,k)\,;\,\,h\in H\,,\,k\in K\}$, then for ANY $\displaystyle h\in H\,,\,\pi_i(h,1)=h$ and we have surjectivity.

Tonio

3. Originally Posted by tonio
I'm guessing you must prove $\displaystyle \pi_1\,,\,\pi_2$ are surjective. Take the first one: since $\displaystyle H\times K=\{(h,k)\,;\,\,h\in H\,,\,k\in K\}$, then for ANY $\displaystyle h\in H\,,\,\pi_i(h,1)=h$ and we have surjectivity.

Tonio
Ok.

So,

Let $\displaystyle \,h, h', e\in H\,,\,k, k', e'\in K\$

(hh',k) = hh' (def )
= h $\displaystyle x$ h'
= (h,k) $\displaystyle x$ (h',k) (def operation H x K)

Since (hh',k) = (h,k) $\displaystyle x$ (h',k) is a homomorphism

Since a homomorphism let $\displaystyle \,e\in H\,,\,e'\in K\$ be the identities.

(h,e) = h (def )

Therefore, a surjective homomorphism.

By a similar argument, $\displaystyle \pi_2$ is also a surjective homomorphism.

QED

This feels very wrong to me...but rarely do proofs ever feel right.

Am I on the right track?

4. Originally Posted by kpizle
Ok.

So,

Let $\displaystyle \,h, h', e\in H\,,\,k, k', e'\in K\$

(hh',k) = hh' (def )
= h $\displaystyle x$ h'
= (h,k) $\displaystyle x$ (h',k) (def operation H x K)

Since (hh',k) = (h,k) $\displaystyle x$ (h',k) is a homomorphism

Since a homomorphism let $\displaystyle \,e\in H\,,\,e'\in K\$ be the identities.

(h,e) = h (def )

Therefore, a surjective homomorphism.

By a similar argument, $\displaystyle \pi_2$ is also a surjective homomorphism.

QED

This feels very wrong to me...but rarely do proofs ever feel right.

Am I on the right track?

$\displaystyle \pi_1\left((h_1,k_1)(h_2,k_2)\right)=\pi_1(h_1h_2, k_1k_2):=h_1h_2=\pi_1(h_1,k_1)\pi_1(h_2,k_2)$

This proves the maps are homomorphisms,a nd you already have surjectivity, so you're done.

Tonio