Kernel of a homomorphism into an abelian group

A group $𝐺$ is abelian iff for any group $𝐻$ and any homomorphism $𝜑 \in 𝖦 𝗋 𝗉 (𝐻, 𝐺)$ the kernel contains the commutator subgroup, i.e.

[𝐻, 𝐻] ⊴ k e r 𝜑

Proof

Let $𝐻$ and $𝐺$ be groups with $𝐻$ abelian and let $𝜑 \in 𝖦 𝗋 𝗉 (𝐺, 𝐻)$ . Then $𝜑 ([𝑎, 𝑏]) = 𝜑 (𝑎) 𝜑 (𝑏) 𝜑 (𝑎) - 1 𝜑 (𝑏) - 1 = 𝑒$ . But every element of the Commutator subgroup $[𝐺, 𝐺]$ is a product of such commutators, so $𝜑 [𝐺, 𝐺] = {𝑒}$ .

Now let $𝐻$ a group such that $[𝐺, 𝐺] \subseteq k e r 𝜑$ for every group $𝐺$ and every homomorphism $𝜑 \in 𝖦 𝗋 𝗉 (𝐺, 𝐻)$ . Let $𝐺 = 𝐻$ and $𝜑 = i d 𝐺$ . It follows that $[𝐻, 𝐻] = {𝑒}$ . Therefore $𝐻$ is abelian.

Kernel of a homomorphism into an abelian group

See also