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I think that for Abelian groups, there is an explicit formula, but it is not pretty, and probably of limited practicial value in all but the smallest cases. Let the minimal order non-trivial subgroups of $G$ be $M_1,M_2,\ldots, M_s.$ In this case, these are just the subgroups of prime order of $G$ (if trying to generalize this for a general finite group, we would look at the minimal normal subgroups). Let $r$ be the dimension we are interested in. We are looking for all characters of the form $\chi = \lambda_1 + \lambda_2 + \ldots + \lambda_r$ with each $\lambda_i$ linear such that $\chi$ is faithful. If $\chi$ is not faithful, there must a minimal non-trivial subgroup $M$ which is in the kernel of $\lambda_i$ for each $i$. So we need to pick the $\lambda_i$ so that for each $j$, there is at least one $\lambda_i$ whose kernel does not contain $M_j$.

For any choice of $j$, there $|G|- [G:M_j]$ linear characters of $G$ which do not contain $M_j$ in their kernels. Let $I = \{1,2,\ldots,s\}$, and for each subset $J$ of $I$ (including the empty set), let $M_J = \langle M_j : j \in J \rangle$. Notice that $\lambda_i$ contains $M_J$ in its kernel if and only if it contains $M_j$ in its kernel for each $j \in J$. For each choice of of $J$, then there are $[G:M_J]$ choices for $\lambda_i$ which contain $M_J$ in their kernels. It is almost easier just to write the formula down than explain it.

Can't do the latex right, but there are $\sum_{J \subseteq I} (-1)^{|J|}$ ($[G:M_J]$ choose $r$) choices of $r$-tuple of linear characters such that the kernel of the sum is trivial. We start with all possible $r$-tuples, corresponding to $J = \emptyset$, then for each $j$, we subtract all $r$ tuples where the kernel of each $\lambda$ occurring contains some $M_j$. Then we have to put back $r$ tuples where the kernel of each $\lambda$ contains $M_{J}$ for some subset $J$ of cardinality $2$, etc.

An explicit formula for a general group could be done, but it would get rather messy.
show/hide this revision's text 3 minor typos

I think that for Abelian groups, there is an explicit formula, but it is not pretty, and probably of limited practicial value in all but the smallest cases. Let the minimal order non-trivial subgroups of $G$ be $M_1,M_2,\ldots, M_s.$ In this case, these are just the subgroups of prime order of $G$ (if trying to generalize this for a general finite group, we would look at the minimal normal subgroups). Let $r$ be the dimension we are interested in. We are looking for all characters of the form $\chi = \lambda_1 + \lambda_2 + \ldots + \lambda_r$ with each $\lambda_i$ linear such that $\chi$ is faithful. If $\chi$ is not faithful, there must a minimal non-trivial subgroup $M$ which is in the kernel of $\lambda_i$ for each $i$. So we need to pick the $\lambda_i$ so that for each $j$, there is at least one $\lambda_i$ whose kernel does not contain $M_j$.

For any choice of $j$, there $|G|- [G:M_j]$ linear characters of $G$ which do not contain $M_j$ in their kernels. Let $I = \{1,2,\ldots,s\}$, and for each subset $J$ of $I$ (including the empty set), let $M_J = \langle M_j : j \in J \rangle$. Notice that $\lambda_i$ contains $M_J$ in its kernel if and only if it contains $M_j$ in its kernel for each $j \in J$. For each choice of of $J$, then there are $[G:M_J]$ choices for $\lambda_i$ which contain $M_J$ in their kernels. It is almost easier just to write the formula down than explain it.

Can't do the latex right, but there are $\sum_{J \subseteq I} (-1)^{|J|}$ ($[G:M_J]$ choose $r$) choices of $r$-tuple of linear characters such that the kernel of the sum is trivial. We start with all possible $r$-tuples, corresponding to $J = \emptyset$, then for each $j$, we subtract all $r$ tuples where the character kernel of each $\lambda$ occurring contains some $M_j$. Then we have to put back $r$ tuples where the kernel of each $\lambda$ contains $M_{J}$ for some subset $J$ of cardinality $2$, etc.

An explicit formula for a general group could be done, but it would get rather messy.
show/hide this revision's text 2 minor typos

I think that for Abelian groups, there is an explicit formula, but it is not pretty, and probably of limited practicial value in all but the smallest cases. Let the minimal order non-trivial subgroups of $G$ be $M_1,M_2,\ldots, M_s.$ In this case, these are just the subgroups of prime order of $G$ (if trying to generalize this for a general finite group, we would look at the minimal normal subgroups). Let $r$ be the dimension we are interested in. We are looking for all characters of the form $\chi = \lambda_1 + \lambda_2 + \ldots + \lambda_r$ with each $\lambda_i$ linear such that $\chi$ is faithful. If $\chi$ is not faithful, there must a minimal non-trivial subgroup $M$ which is in the kernel of $\lambda_i$ for each $i$. So we need to pick the $\lambda_i$ so that for each $j$, there is at least one $\lambda_i$ whose kernel does not contain $M_j$.

For any choice of $j$, there $|G|- [G:M_j]$ linear characters of $G$ which do not contain $\lambda_i$M_j$ in their kernels. Let $I = \{1,2,\ldots,s\}$, and for each subset $J$ of $I$ (including the empty set), let $M_J = \langle M_j : j \in J \rangle$. Notice that $\lambda_i$ contains $M_J$ in its kernel if and only if it contains $M_j$ in its kernel for each $j \in J$. For each choice of of $J$, then there are $[G:M_J]$ choices for $\lambda_i$ which contain $M_J$ in their kernels. It is almost easier just to right write the formula down than explain it.

Can't do the latex right, but there are $\sum_{J \subseteq I} (-1)^{|J|}$ ($[G:M_J]$ choose $r$) choices of $r$-tuple of linear characters such that the kernel of the sum is trivial. We start with all possible $r$-tuples, corresponding to $J = \emptyset$, then for each $j$, we subtract all $r$ tuples where the character of each $\lambda$ occurring contains some $M_j$. Then we have to put back $r$ tuples where the kernel of each $\lambda$ contains $M_{J}$ for some subset $J$ of cardinality $2$, etc.

An explicit formula for a general group could be done, but it would get rather messy.
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