Is character values on $G$-modules additive? The following definition is from Atiyah & Macdonald. Let $C$ be a class of $A$-modules, where $A$ is a commutative ring with identity. Let $\lambda$ be a function on $C$ with values in an abelian group $G$. The function $\lambda$ is $\textit{additive}$ if for each short exact sequence $$0\to M' \to M \to M'' \to 0,$$ whose terms are in $C$, we have $\lambda(M')-\lambda(M)+\lambda(M'')=0$. For example, a simple example is let $C$ be the class of all finite-dimensional $k$-vector spaces $V$, then the map $V\mapsto \dim{V}$ is an additive function on $C$.
Now consider $C$ to be the space of $G$-modules/$G$-representations (by definition this is not an $A$-modules in the sense of Atiyah and MacDonald since $G$ is not a ring), if we have a short exact sequence of $G$-modules $$0\to M'\to M\to M'' \to 0,$$
is the character value $\chi(g)$ of an element $g\in G$ addtive on $C$, that is if $\chi_{M'}(g)-\chi_M(g)+\chi_{M''}(g)=0$?
 A: Edit: Fun, but irrelevant fact: a $G$-representation actually is a module in the sense of Atiyah-MacDonald, over a ring called the group algebra (or group ring). It's fun to show that a $G$-representation really is a module and in the ordinary sense, so I'd recommend giving it a go! https://en.wikipedia.org/wiki/Group_ring
We will make extensive use of the fact that representations are $k$-vector spaces, and basically translate this into a linear algebra problem.  By basic properties of short exact sequences, we may assume that $M' \subseteq M$ and $M'' = M/M'$ is the quotient representation.
Let $\{v_1, \ldots, v_m\}$ be a basis for $M'$ and extend it to a basis $\{v_1,\ldots, v_n\}$ for $M$.
Let $T:M \to M$ be the linear transformation corresponding to the action of $g$. Then $g$ acts on $M'$ by the restriction $T\mid_{M'}$ and $g$ acts on $M'' = M/M'$ by the induced map $\bar{T}$. Let $T$ have matrix $A = (a_{ij})_{1\leq i, j \leq n}$ with respect to the basis $\{v_1,\ldots, v_n\}$.
The value $\chi_{M'}(g)$ is the trace of the linear transformation $T\mid_{M'}:M' \to M'$. Since $\{v_1,\ldots, v_m\}$ is a basis for $M'$, it is easy to see that $\chi_{M'}(g) = \operatorname{tr}(\bar{T}) = \sum_{i=1}^m a_{ii}$.
We have that $\{v_{m+1} + M', \ldots, v_{n} + M'\}$ is a basis for $M'' = M/M'$, and $\bar{T}$ acts on this basis by
$$
T(v_j + M') = \sum_{i=1}^n a_{ij}(v_i + M') = \sum_{i=m+1}^na_{ij}(v_i + M'),
$$
which means that the diagonal entries of the matrix of $\bar{T}$ are $a_{ii}$ for $m + 1 \leq i \leq n$, and hence the trace of this map is $\sum_{i=m+1}^n a_{ii}$, so $\chi_{M''}(g) = \sum_{i=m+1}^n a_{ii}$.
Clearly $\chi_{M}(g) = \sum_{i=1}^n a_{ii}$, so we have
$$
\chi_{M}(g) = \sum_{i=1}^n a_{ii} = \sum_{i=1}^m a_{ii} + \sum_{i=m+1}^n a_{ii} = \chi_{M'}(g) + \chi_{M''}(g).
$$
