Any morphism $\phi:G \rightarrow A$ to an abelian group $A$ factors uniquely through the projection $G \rightarrow G /[G,G]$.

abelian-groupsabstract-algebragroup-homomorphismgroup-theorynormal-subgroups

I'm doing this exercise 7(b) in textbook Algebra by Saunders MacLane and Garrett Birkhoff. Could you please verify if my attempt is fine or contains logical mistakes?

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Let $G$ be a group and $C$ its commutator subgroup. Prove that

  • $G / C$ is abelian.

  • Any morphism $\phi:G \rightarrow A$ to an abelian group $A$ factors uniquely through the projection $G \rightarrow G /C$.

My attempt:

For $a,b \in G$, we have $aC, bC \in G/C$. It follows from $b^{-1}a^{-1}ba \in C$ that $C =
(b^{-1}a^{-1}ba)C$. Then $(aC)(bC) = (ab)C = (ab)(b^{-1}a^{-1}ba)C=(ba)C = (bC)(aC)$. Hence $G/C$ is abelian.

Next we prove that $\phi(C) = \{1\}$. For $x = b^{-1}a^{-1}ba \in C$, we have $\phi(x) = \phi(b^{-1}a^{-1}ba) = \phi(b)^{-1} \phi(a)^{-1} \phi(b) \phi(a)$. On the other hand, $A$ is abelian and thus $\phi(a)^{-1} \phi(b) = \phi(b) \phi(a)^{-1}$. Hence $\phi(x) = 1$.

To sum up, we have $C \trianglelefteq G$ and $\phi:G \rightarrow A$ a group morphism and $\phi(C) = \{1\}$. Then the result follows from Theorem 26.

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Best Answer

What you have looks great! In the interest of precision, you might mention that $C$ is generated by the commutators $[a,b]$. Because of this, showing that $\phi([a,b]) = 1$ for all $a,b$ (as you did) is enough to show that $\phi_* C = \{1\}$ (as is required). This is a subtle point, though, and as long as you know that it works, that's enough.

I hope this helps ^_^

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