Commutator subgroup is the minimal normal subgroup such that quotient group is abelian

Asked Nov 02 '15 at 09:44

Active Nov 02 '15 at 09:51

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I was recently asked this in Abstract Algebra class on group theory:

Let G be a group and G' its commutator subgroup (i.e. its minimal subgroup containing all commutators $ [x,y] = xyx^{-1}y^{-1} $ where $ x,y \in G $ ). Where are asked to show G' is the minimal normal subgroup such that the quotient group $ G/G' $ is Abelian (if $ H \unlhd G $ and G/H is Abelian then we must have $ G' < H $)

To be honest I have tried but could not seem to be able to solve this. I have shown G' is normal and quotient group $ G/G' $ is abelian, but I cannot prove G' is minimal normal subgroup with this property. I really would appreciate the help on this. Thanks all

edited Nov 02 '15 at 09:51

asked Nov 02 '15 at 09:44

Don John Prep

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Which part is the problem? Showing that the quotient is abelian or that $G'$ is minimal with this property? – Tobias Kildetoft Nov 02 '15 at 09:47
@TobiasKildetoft : it is showing that $ G' $ is minimal with this property of all normal subgroups, I have shown quotient group is abelian – Don John Prep Nov 02 '15 at 09:50
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So take some normal subgroup $N$ and assume that $G/N$ is abelian. Write up what this means in terms of commutators. – Tobias Kildetoft Nov 02 '15 at 09:51
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@AbeerAbassi If $\phi\colon G\to A$ is a homomorphism and $A$ is abelian, what can you say about $\phi(xyx^{-1}y^{-1})$? – Hagen von Eitzen Nov 02 '15 at 09:51

Commutator subgroup is the minimal normal subgroup such that quotient group is abelian

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