For a general Hopf algebra $H$, the coproduct of the identity element is $$\Delta(1) = 1\otimes 1.$$ Now for a finite group $G$ and a field $k$, $k(G)$ forms a Hopf algebra with basis $\{\delta_{g}| g\in G\}$. The coproduct is given $$\Delta(\delta_{g}) = \sum_{g_{1}g_{2} = g}\delta_{g_{1}}\otimes\delta_{g_{2}}.$$ The identity element $1$ is $$1 = \sum_{g\in G}\delta_{g}.$$ I want to establish that $\Delta(1) = 1\otimes 1$. This is what I have done,
$$\Delta(1) = \Delta\left(\sum_{g\in G}\delta_{g}\right) = \sum_{g\in G}\Delta(\delta_{g}) = \sum_{g\in G}\left(\sum_{g_{1}g_{2} = g}\delta_{g_{1}}\otimes\delta_{g_{2}}\right).$$ How do I proceed from here?
I am thinking I can use this change of indices but I am not too sure if it is correct $$\sum_{g\in G}\sum_{g_1g_2 = G} = \sum_{g_2\in G}\sum_{g_1 = gg_2^{-1}}$$
Observe $\delta_{g_1}\otimes\delta_{g_2}$ appears exactly once for every possible pair $(g_1,g_2)$, then we may factor:
$$ \sum_{g\in G} \Big(\sum_{g_1g_2=g} \delta_{g_1}\otimes\delta_{g_2}\Big)=\sum_{g_1\in G}\sum_{g_2\in G} \delta_{g_1}\otimes\delta_{g_2}=\Big(\sum_{g_1\in G}\delta_{g_1}\Big)\otimes\Big(\sum_{g_2\in G} \delta_{g_2}\Big)=1\otimes1. $$