# External direct product of a family of group homomorphisms

The following is from Hungerford's Algebra book:

Theorem 8.10. Let $$\{f_i\colon G_i\to H_i\mid i\in I\}$$ be a family of homomorphisms of groups and let $$f=\prod f_i$$ be the map $$\prod\limits_{i\in I}G_i \to \prod\limits_{i\in I}H_i$$, given by $$\{a_i\}\longmapsto \{f_i(a_i)\}$$. Then $$f$$ is a homomorphism of groups such that $$f\left({\prod\limits_{i\in I}}^w G_i\right)\subset {\prod\limits_{i\in I}}^w H_i$$, $$\ker f = \prod\limits_{i\in I}\ker f_i$$, and $$\mathrm{Im}\,f = \prod\limits_{i\in i}f_i$$. Consequently, $$f$$ is a monomorphism [resp. epimorphism] if and only if each $$f_i$$ is.

The only part that I am struggling with is that $$f$$ is a homomorphism of groups, which I don't think to be true, since for that each $$H_i$$ needs to be Abelian but the theorem doesn't mention that. So how is $$f$$ a group homomorphism?

An element $$\mathbf{x}=(x_i)_{i\in I}$$ of a product $$\prod K_i$$ is completely determined by its "components", that is, the values $$\pi_j(\mathbf{x})=x_j$$. We know that $$\mathbf{x}=\mathbf{y}$$ if and only if $$\pi_j(\mathbf{x}) = \pi_j(\mathbf{y})$$ for all indices $$j$$.
Now, if $$(a_i),(b_i)\in \prod G_i$$, then their product is define "pointwise", that is, $$\pi_j\Bigl( (a_i)(b_i)\Bigr) = \pi_j(a_i)\pi_j(b_i) = a_jb_j.$$
The definition of $$\prod f_i$$ is that $$\pi_j\left(\left(\prod f_i\right)(x_k)\right) = f_j\Bigl(\pi_j(x_k)\Bigr) = f_j(x_j).$$
Thus, we have: \begin{align*} \pi_j\left(\left(\prod f_i\right)\bigl((a_i)(b_i)\bigr)\right) &= f_j\Bigl(\pi_j\bigl((a_i)(b_i)\bigr)\Bigr)\\ &= f_j\Bigl(\pi_j(a_i)\pi_j(b_i)\Bigr)\\ &= f_j\Bigl(a_jb_j\Bigr)\\ &= f_j(a_j)f_j(b_j).\\ \pi_j\left(\left( \left(\prod f_i\right)(a_i)\right)\left(\left(\prod f_i\right)(b_i)\right)\right) &=\pi_j\left( \left(\prod f_i\right)(a_i)\right) \pi_j\left(\left(\prod f_i\right)(b_i)\right)\\ &= f_j(\pi_j(a_i))f_j(\pi_j(b_i))\\ &= f_j(a_j)f_j(b_j). \end{align*}
Therefore, $$\left(\prod f_i\right)\Bigl( (a_i)(b_i)\Bigr) = \left( \left(\prod f_i\right)(a_i)\right) \left( \left(\prod f_i\right)(b_i)\right),$$ proving that we have a homomorphism.