Given a subgroup H, show there are distinct characters which are same on H. 
Given a subgroup $H\subset G$ and a conjugacy class $C$ in $G$ disjoint from $H$, show
that there are distinct characters $\chi_1,\chi_2$ that agree on $H$.

Somebody else answered this question (see their corollary), but I still don't see the proof. They argue that one can define class functions $f,g$ that only differ in their values on $C$, for example
$$f(C)=1$$
$$ g(C)=0 $$
$$ f|(G-C)=g|(G-C)=1 $$
and they argue that would mean there are characters $\phi_1,\phi_2$ which agree on $H$ but not everywhere.
One can indeed define these class functions $f,g$. But these might be an arbitrary linear combination of the irreducible characters $\chi_1,\dots,\chi_n$ of $G$, like
$$f = \sum_{i=1}^n a_i \chi_i$$
$$g = \sum_{i=1}^n b_i \chi_i$$
where $a_i,b_i$ are not necessarily rational numbers, so one cannot multiply by a large enough integer and get integer coefficients to get characters, which are defined to lie on non-negative integer span of irreducible characters, i.e. $\bigoplus_{i=1}^n\mathbb{N_0}\chi_i$.
So how would one get characters which are same on $H$ but differ on (presumably) $C$?
 A: Heres my try for finite groups $G$: Showing the statement for class functions is much easier than showing it for virtual characters, let alone actual characters. However, we can use a similar idea of showing that $\dim(V_{\rm irr})$ divides $|G|$ where we show that coefficients exist both in $\overline{\mathbb{Z}}$ and $\mathbb{Q}$. Enumerating the conjugacy classes, we get $C_1,...,C_k,C_{k+1},...,C_d$ where $C_1,...,C_k$ have a nonempty intersection with $H$ and $C_{k+1},...,C_d$ are disjoint from $H$. By assumption, $k < d$. Let $\chi_1,...,\chi_d$ be the characters of the irreps. With an orthonormal basis, $\rho(g)$ has roots of unity on the diagonals, so $\chi(g) \in \overline{\mathbb{Z}}$, i.e., is an algebraic integer. Consider the $d$ vectors $v_i = (\chi_i(C_1),...,\chi_i(C_k))$, corresponding to the character of irrep $V_i$ on the first $k$ conjugacy classes. Viewing the vectors over $\overline{\mathbb{Q}}$, since $k < d$, basic linalg dictates that there exist $a_i \in \overline{\mathbb{Q}}$ not all zero such that $a_1 v_1 + ... + a_dv_d = 0$. Thus, the character $a_1\chi_1 + ... + a_d \chi_d$ is $0$ on $C_1,...,C_k$ and therefore $0$ on $H$.
We can now leverage this solution over $\overline{\mathbb{Q}}$ to get a solution over $\mathbb{Q}$. Let $\chi_1',...,\chi_r'$ be the irreducible characters of $H$. Note that $r$ might be a lot larger than $d$ — just because $H$ is a subgroup of $G$ does not mean it has fewer conjugacy classes. It is easy to check that each $\chi|_H$ is still a class function on $H$. Thus, each $\chi_i|_H$ is an integral sum of the $\chi_j'$. So there exist $b_i = (b_{i,1},...,b_{i,r})$ such that $\chi_i|_H = b_{i,1}\chi_1' + ... + b_{i,r}\chi_r'$. If $r$ was less than $d$, then we could make another linalg argument to say that there must exist nonzero rationals $q_i$ such that $q_1 b_1 + ... + q_d b_d = 0$, and we would be done. However, $r$ might be larger than $d$. We instead use the $a_i$ from the previous paragraph, getting that $a_1b_1 + ... + a_d b_d$ corresponds to the zero character on $H$. Since the $\chi_i'$ are an orthonormal basis, it follows that $a_1 b_1 + ... + a_d b_d = (0,...,0)$. Thus, we have the matrix equation
$$\begin{pmatrix} 
b_{11} & \cdots & b_{d1} \\
\vdots & \ddots & \vdots \\
b_{1r} & \cdots & b_{dr} \\
\end{pmatrix}\begin{pmatrix} 
a_1 \\
\vdots  \\
a_d \\
\end{pmatrix} = \textbf{0},$$ where $b_{i,j} \in \mathbb{Z}^{\geq 0}$ and $a_i \in \overline{\mathbb{Q}}$ and not all zero. It is also important to note that each $b_i$ is nonzero since each $\chi_i|_H \neq 0$. As mentioned, if $r < d$ we are done. Otherwise, we can choose $d$ rows of $B$ with at least one corresponding $a_i$ in nonzero and the resulting matrix $B'$ will have determinant zero. Thus, we eventually can find a rational solution $a' \in \mathbb{Q}^r$ such that $Ba' = 0$. Essentially, we use the vector $a \in \overline{\mathbb{Q}}^d$ and the fact that all the $b_{ij}$ are integers to show that there must be some additional vector solution $a'$ in the fraction field of $\mathbb{Z}$. By scaling the denominators, this gets us some integers $c_i$ with at least one nonzero such that $c_1\chi_1 + ... + c_d\chi_d$ is zero on $H$. Since one $c_i$ is nonzero, $c_1\chi_1 + ... + c_d\chi_d$ is not zero on $G$ by the Fundamental Theorem of Character Theory. Then, we just need to add a large enough character $\chi$ such that all the coefficients are positive. Then, $\chi$ and $\chi + c_1\chi_1 + ... + c_d\chi_d$ will match on $H$ but will differ on $G$, proving the existence of such characters.
Let me know what you think! And please correct any mistakes.
