# Commutator Subgroup and the Largest Abelian Quotient

Good day. I'm not too sure how I should go about asking my question, so if it is in anyway confusing, feel free to edit it.

In my Group Theory course we defined and showed a few properties of the commutator subgroup $[G,G]$ of a group $G$. And then we stated and proved a lemma in lecture that I'm having some difficulty wrapping my head around. I'll state it here for clarity.

Lemma
a) The commutator subgroup $[G,G]$ of $G$ is a normal subgroup of $G$.
b) The quotient group $G / [G,G]$ is the largest abelian quotient of G in the sense that every homomorphism $\varphi : G \to A$ where $A$ is an an abelian group, induces a unique homomorphism $\bar \varphi : G/[G,G] \to A$ so that $\varphi =\bar \varphi \circ \pi$ where $\pi : G \to G/[G,G]$ is the quotient homomorphism.

Now, it's not the proof of this lemma that I'm having trouble with, it's the italicized text that is messing with me. I completely understand what it is that the italicized text is saying, I'm just not too sure why it is that the property gives us the "largeness" that is claimed. I can see that the above forces $G$ to "pass through" $G/[G,G]$ before it "reaches" $A$ (I apologize for the casual terminology), so I can sort of see why $G/[G,G]$ should be the largest quotient, in a non-rigorous sort of way.

Overall I'm very confused. Perhaps I'm missing something very simple. Any help or nudges in the right direction would be much appreciated.

As a side note I did try a search on Google and here on Stack Exchange and couldn't find anything that answered my question. So if I missed something obvious, then my apologies. Thank you in advance.

## 1 Answer

Remember that the larger $N\lhd G$ is $G$ is, the smaller $G/N$ will be. Now suppose $N\lhd G$ and $G/N$ is abelian. This means that $abN=baN$ for any $a,b\in G$. This translates in $a^{-1}b^{-1}abN=N$. This means $N$ contains all commutators, and hence contains $[G,G]$. Thus, whenever $N$ is normal and the quotient $G/N$ is abelian $[G,G]\leq N$. Thus $G/[G,G]$ is as big as $G/N$ gets whenever $N$ is normal and $G/N$ is abelian.

• Thank you. That really helped actually. Correct me if I'm wrong, but essentially the line of reasoning is (supposing N is normal): G/N abelian implies N contains the commutator subgroup, and since the commutator subgroup is normal and has order less than or equal to the order of N, G/[G,G] must have order that is bigger than or equal to G/N, hence we get that G/[G,G] is the largest abelian quotient? – QuantumLeap Dec 12 '13 at 4:14
• @CanadianStudent Aha. – Pedro Tamaroff Dec 12 '13 at 4:17
• Well thank you very much! I figured there was something simple I was missing. – QuantumLeap Dec 12 '13 at 4:19