1. ## Homomorphism question.

I think I'm confused about notation perhaps.

let $p: \mathbb{Z} x \mathbb{Z} -> G$ be a homomorphism

define $p(m,n)=h^{m}k^{n}$ for h,k in G

give a necessary and sufficient condition for it to be a homomorphism... prove why.

i know to show something is a homomorphism i have to show that $p(ab)=p(a)p(b)$, ie. that the binary operation's action is preserved in both groups. but i don't know how to here. for a=(m,n) and b=(i,j), is ab=(mi,nj)? that would give $h^{mi}k^{nj}=h^{m+i}k^{n+j}$. How do I go about setting conditions to make these equal...? Seems to me they only would be if h,k were 0 or 1?

2. Originally Posted by platinumpimp68plus1
I think I'm confused about notation perhaps.

let $p: \mathbb{Z} x \mathbb{Z} -> G$ be a homomorphism

define $p(m,n)=h^{m}k^{n}$ for h,k in G

give a necessary and sufficient condition for it to be a homomorphism... prove why.

i know to show something is a homomorphism i have to show that $p(ab)=p(a)p(b)$, ie. that the binary operation's action is preserved in both groups. but i don't know how to here. for a=(m,n) and b=(i,j), is ab=(mi,nj)? that would give $h^{mi}k^{nj}=h^{m+i}k^{n+j}$. How do I go about setting conditions to make these equal...? Seems to me they only would be if h,k were 0 or 1?
I think it must be $\mathbb{Z}\oplus\mathbb{Z}$ , which sometimes is simply written $\mathbb{Z}\times\mathbb{Z}$ = the free abelian group of rank 2.

The operation here's defined as $(m,n)+(m',n')=(m+m',n+n')$ , so your function's a homom. iff $p(m+m',n+n')=p(m,n)+p(m',n')\Longleftrightarrow\,\ ,h^{m+m'}k^{n+n'}=(h^m k^n)(h^{m'} k^{n'})$

Can you see now what the condition on G has to be?

Tonio

3. Ooh okay, I always get confused about notation.

So G has to be abelian then, ie. hk=kh ?

4. Originally Posted by platinumpimp68plus1
Ooh okay, I always get confused about notation.

So G has to be abelian then, ie. hk=kh ?

It seems so, but the way you defined h doesn't make it clear: what are those two elements in G? Generators or what? Or h, k are only two arbitrary elements in G? If the latter then all it's requires is that h, k commute, though G itself can be non-abelian.

Tonio

5. Originally Posted by tonio
It seems so, but the way you defined h doesn't make it clear: what are those two elements in G? Generators or what? Or h, k are only two arbitrary elements in G? If the latter then all it's requires is that h, k commute, though G itself can be non-abelian.

Tonio

they were just defined as arbitrary elements in G.