proving surjective homomorphism

**kpizle** · November 29th 2009, 02:59 PM

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?

Please help!

Thanks.

**tonio** · November 29th 2009, 03:55 PM

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?

Please help!

Thanks.

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

Tonio

**kpizle** · November 29th 2009, 04:42 PM

Originally Posted by tonio

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

Tonio

Ok.

So,

Let $<br /> \,h, h', e\in H\,,\,k, k', e'\in K\$

(hh',k) = hh' (def

)
= h $x$ h'
=

(h,k) $x$

(h',k) (def operation H x K)

Since

(hh',k) =

(h,k) $x$

(h',k) is a homomorphism

Since

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

(h,e) = h (def

)

Therefore,

a surjective homomorphism.

By a similar argument, $\pi_2$ 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?

**tonio** · November 29th 2009, 06:46 PM

Originally Posted by kpizle

Ok.

So,

Let $<br /> \,h, h', e\in H\,,\,k, k', e'\in K\$

(hh',k) = hh' (def

)
= h $x$ h'
=

(h,k) $x$

(h',k) (def operation H x K)

Since

(hh',k) =

(h,k) $x$

(h',k) is a homomorphism

Since

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

(h,e) = h (def

)

Therefore,

a surjective homomorphism.

By a similar argument, $\pi_2$ 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?

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

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

Tonio

**kpizle** · November 29th 2009, 07:04 PM

Ooohhh. The missing link!

Thank you very much.

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