Kernel of a homomorphism of a group algebra Excuse me, this is stupid, but I have a short circuit in my head, I can't understand the situation.
Let $k$ be a field, $G$ a finite group,  $kG$ the corresponding group algebra and $\delta:G\to kG$ the natural embedding. Let $A$ be a unital associative algebra over $k$ and $\pi:G\to A$ a homomorphism, i.e. a map with the following properties:
$$
\pi(a\cdot b)=\pi(a)\cdot \pi(b),\qquad \pi(1)=1,\qquad a,b\in G.
$$
Then there is a homomorphism of algebras $\varphi:kG\to A$ such that
$$
\varphi\circ\delta=\pi.
$$
Consider the kernels of $\pi$ and $\varphi$:
$$
{\tt Ker}(\pi)=\{a\in G:\ \pi(a)=1\},\qquad {\tt Ker}(\varphi)=\{x\in kG:\ \varphi(x)=0\}.
$$
A question:

What is the connection between ${\tt Ker}(\pi)$ and ${\tt Ker}(\varphi)$? Is it possible  to express ${\tt Ker}(\varphi)$ in terms of ${\tt Ker}(\pi)$, and vice versa?

 A: I don't think there necessarily is a very clear connection. Consider the following example for starters. Let $k=\Bbb{C}$ and $G=C_n=\langle g\rangle$ a cyclic group. Furthermore, let $A=\Bbb{C}$. Then we can for example compare $\pi_0$, the trivial homomorphism, $\pi_0(g^k)=1$ for all $k$, and the injective homomorphism $\pi_1(g^k)=e^{2\pi i k/n}$. The kernels of the two homomorphisms $\pi_0,\pi_1:G\to A$ are as far apart as can be. Yet the dimensions of the kernels of the lifts $\phi_0,\phi_1:kG\to A$ both have dimension $n-1$.
The above generalizes as follows in a situation where Maschke's theorem applies, i.e. $|G|\cdot1_k\neq0_k$. If $V$ is any irreducible module, then associated with it we have a homomorphism $\phi_V:kG\to GL(V)$. Here we can choose $A=GL(V)$ and, of course, we get a matching $\pi_V$ by restricing this to $G$. As $GL(V)$ is one of the Wedderburn components of $kG$, the kernel of $\phi_V$ is the sum of the other Wedderburn components. OTOH $g\in\ker\pi_V$ iff $\chi_V(g)=\dim V$.
Going in the other direction we can consider the $k$-span $M$ of $\pi(G)$. This is clearly a cyclic $kG$-module generated by $\pi(1_G)$, and also a subalgebra of $A$, because multiplicativity of $\pi$ means that it is closed under multiplication. So we can replace $A$ with $M$, when $\phi$ is surjective. The kernel of $\phi$ is a 2-sided ideal of $kG$, so necessarily a sum of Wedderburn components (again assuming that $k=\Bbb{C}$).
A: The following is not right in general (see comments), but may be helpful to others looking for a solution.
$\pi$ is a homomorphism (of monoids with unit) from $G$ into the multiplicative monoid of $A$. Since $\pi(a) \cdot \pi(a^{-1}) = \pi(a \cdot a^{-1}) = \pi(1) = 1$, the image of $\pi$ is actually a subgroup, $H$ say, of the multiplicative monoid of $A$. The subalgebra of $A$ generated by $H$ will then be isomorphic to the group algebra $kH$. Let $m = |\ker(\pi)|$ and $n = |H|$, then we have
$$
|G| = mn
$$
So as $kG$ is $|G|$-dimensional and $kH$ is $m$-dimensional, we have
$$
\dim(\ker(\phi)) = |G| - m = (m - 1)n
$$
If you enumerate the non-unit elements of $\ker(\pi)$ as $k_1, k_2, \ldots, k_{m-1}$ and pick representatives $g_1, g_2, \ldots, g_n$ for the cosets of $\ker(\pi)$, then the elements $k_ig_j - g_j$ of $kG$ give you $(m-1)n$ linearly independent elements of $\ker(\phi)$ and hence form a basis for $\ker(\phi)$.
