It is known that the category of commutative and cocommutative Hopf algebras over a field $k$ is an abelian category. So we can talk about exact sequences. Reading a paper I found the following reasoning: if we have an exact sequence of such Hopf algebras \begin{equation} k \rightarrow H_1 \rightarrow H_2 \rightarrow H_3 \rightarrow \dots \rightarrow H_n \rightarrow k \end{equation} and all the involved algebras are finite dimensional when considered as $k$-vector spaces then we have $\prod_i \dim_kH_{2i}=\prod_{j}\dim_kH_{2j+1}$. This is different from the context of $k$-vector spaces where the alternating sums of dimensions coincide.
I considered as starting example the following: take $N \trianglelefteq G$ with $G$ finite abelian group, then \begin{equation} k \rightarrow k[N] \rightarrow k[G]\rightarrow k[G/N] \rightarrow k \end{equation} is a short exact sequence and clearly $\dim_k k[G]=|G|=|H||G/H|=\dim_k k[H] \cdot \dim_kk[G/H]$. So the above formula seems plausible.
But I have trouble proving it. It is enough to show it for a short exact sequence \begin{equation} k \rightarrow A \rightarrow B \rightarrow C \rightarrow k. \end{equation} My idea is the following: we fix a basis $a_1, \dots, a_k$ for $A$ and $c_1, \dots, c_l$ for $C$ such that $a_1$ and $c_1$ are the units of $A$ and $C$ respectively. Then we take $d_1, \dots d_l \in B$ lifting the $c_i$'s, I want to show $\{ a_i d_j\}_{i,j}$ forms a basis for $B$.
I am stuck on the linear independence. If we have a combination $\sum \lambda_{ij}a_i d_j=0$ then we can apply $q$ and using the fact that $q(a_1)=c_1$ and $q(a_i)=0$ for $i>1$ we get $\sum_{j}\lambda_{1j}c_j=0$ hence $\lambda_{1j}=0$ for all $j$. But I do not see how I can prove the other coefficients are $0$. If $a_i$ would be invertible we could multiply the linear combination by $i(a_i^{-1})$ and apply again $q$ to get $\sum_j \lambda_{ij}c_j=0$. But in full generality $a_j$ has not this property.
Any idea about how to complete the proof?