The following can be found in Montgomery, Hopf Algebras and their Actions on Rings, example 1.3.2.
Let $G$ be a group and $B=kG$ be its groups algebra, where $k$ is a field. Then $B$ is a bialgebra via $\Delta g = g \otimes g$ and $\epsilon(g)=1$ for all $g \in G$.
I am trying to prove this. It's clear to me that multiplication $m$ and scalar multiplication $u$ make $B$ to an algebra, and the defined comultiplication and counit make $B$ into a coalgebra. I want to show that $m$ and $u$ are coalgebra homomorphisms. More precisely, I want to show that $\Delta_B \circ m = (m \otimes m) \circ \Delta_{B\otimes B}$ and $\epsilon_B \circ m = \epsilon_{B \otimes B}$, and the same for $u$.
For the first assertion, let $h \in G$ and $\sum_{g\in G}\lambda_g g \otimes \sum_{g\in G}\mu_g g \in B \otimes B$. Now compute $$ \begin{split} \Delta_B \circ m (\sum_{g\in G}\lambda_g g \otimes \sum_{g\in G}\mu_g g) (h)&= \Delta_B (\sum_{g \in G}(\sum_{ab=h} \lambda_a \mu_b) g)\\ &=\sum_{g \in G}(\sum_{ab=h} \lambda_a \mu_b) g) \otimes \sum_{g \in G}(\sum_{ab=h} \lambda_a \mu_b) g)\\ &= (m \otimes m) (\sum_{g\in G}\lambda_g g \otimes \sum_{g\in G}\mu_g g) \otimes (\sum_{g\in G}\lambda_g g \otimes \sum_{g\in G}\mu_g g) (h)\\ &= (m \otimes m) \circ \Delta_{B \otimes B} (\sum_{g\in G}\lambda_g g \otimes \sum_{g\in G}\mu_g g) \end{split}$$
Now when I attempt to prove the second identity, I get: $$ \begin{split} \epsilon_B \circ m (\sum_{g\in G}\lambda_g g \otimes \sum_{g\in G}\mu_g g) (h)&= \epsilon_B (\sum_{g \in G}(\sum_{ab=h} \lambda_a \mu_b) g)\\ &=\sum_{ab=h} \lambda_a \mu_b \end{split} $$ Now, how do I evaluate $\epsilon_{B\otimes B}(\sum_{g\in G}\lambda_g g \otimes \sum_{g\in G}\mu_g g) (h)$?