# Homomorphism

• Jan 27th 2009, 05:21 PM
Mel
Homomorphism
Totally lost again!
• Jan 28th 2009, 10:54 AM
clic-clac
That's algebra! :)

$\displaystyle \forall a\in G_1,\forall (x,y)\in G,\ (x,y)(a,e_2)(x,y)^{-1}=(x,y)(a,e_2)(x^{-1},y^{-1})$ $\displaystyle =(xax^{-1},ye_2y^{-1})=(xax^{-1},e_2)$
$\displaystyle a,x\in G_1 \Rightarrow xax^{-1}\in G_1 \Rightarrow (xax^{-1},e_2)\in N$
Therefore $\displaystyle N$ is a normal subgroup of $\displaystyle G$.

$\displaystyle h:G_1\rightarrow N:x\mapsto (x,e_2)$ is surjective by definition of $\displaystyle N$ , a group homomorphism because $\displaystyle h(xy)=(xy,e_2)=(x,e_2)(y,e_2)=h(x)h(y)$ and injective because $\displaystyle (x,e_2)=(e_1,e_2)\Rightarrow x=e_1$ .
We have found an isomorphism between $\displaystyle G_1$ and $\displaystyle N$.

Consider $\displaystyle \phi: G\rightarrow G_2:(a,b)\mapsto b$ .
$\displaystyle \phi((a,b)(c,d))=\phi((ac,bd))=bd=\phi((a,b))\phi( (c,d))$ , so $\displaystyle \phi$ is an homomorphism, which is surjective by definition of $\displaystyle G$.
Moreover, $\displaystyle \forall (a,b)\in G,\ \phi((a,b))=e_2\Leftrightarrow b=e_2 \Leftrightarrow (a,b)\in N$.
Hence $\displaystyle \ker\phi = N$ , and we can conclude $\displaystyle G/N\cong G_2$ .