# Homomorphism

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

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

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

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