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Also I need to show that every 1-dimensional representation of $G$ arise from some 1-dimensional representation of $G/N$.

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Just to inform, that subgroup is the commutator of $G$, $[G,G]$. – Sigur Oct 20 '12 at 22:05
I've changed algebra tag to abstract-algebra, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Oct 21 '12 at 7:08
up vote 3 down vote accepted

To show that $N\lhd G$, it suffices to show $gng^{-1}\in N$ where $n\in N$ and $g\in G$

Hint: Try to write down an element $n\in N$. What does it look like?
Then, try to convert $gng^{-1}$ into a form that is in $N$.

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Without many words (this is assumed to be only a hint as this questions smells like homework). Let us write $\,xyx^{-1}=y^x\,$ for simplicity (this is also a widespread notation):


Finally, check that in general $\,(y^{-1})^x=(y^x)^{-1}\,$

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The subgroup of $G$ generated by all commutators $ghg^{-1} h^{-1}$ is obviously invariant under every automorphism of $G$, i.e. it is a characteristic subgroup of $G$. And in general, characteristic subgroups are normal (use inner automorphisms).

As already noted in the comments, the subgroup generated by all commutators is known as the commutator subgroup $[G,G]$. Clearly, $G^{ab} := G/[G,G]$ is abelian, and the homomorphism theorem immediately implies the following universal property: If $A$ is an abelian group, then every homomorphism $G \to A$ factors uniquely through the projection $G \to G^{ab}$. In order words, there is a natural bijection $\hom(G^{ab},A) \cong \hom(G,A)$. Now apply this to $A=\mathbb{C}^*=\mathrm{GL}(1,\mathbb{C})$. Then this exactly says that $1$-dimensional representations of $G$ correspond to $1$-dimensional representations of $G^{ab}$.

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