Prove that a subset $\{a \in A \mid \chi (a) = 1\}$ is an $(n - 1)$-dimensional subspace of $V$. 
Let $A$ be the additive group of n-dimensional vector space  $V$ over the field $\mathbb{F}_p$. Let $\chi$ - a nontrivial irreducible complex character of A. Prove that a subset $\{a \in A \mid \chi (a) = 1\}$ is an $(n - 1)$-dimensional subspace of $V$.

There are hints in my book. Use the following statement:
1)Let $\Phi$ be $n$-dimensional complex representation of a finite group $G$. $$\chi (g) = n \Leftrightarrow g \in \ker(\Phi).$$
2) Let $A_1$ be a subgroup of $A.$ If $[A:A_1]=p$ then $\exists$ an $(n - 1)$-dimensional subspace $ V_1$ such that $A_1$ be subgroup of elements of $V_1$
I get tripped up on this problem . I proved statement 1. But I can not prove statement 2. I do not see how to apply these hints to the basic problem.
I need to see how this is done step by step.
Sorry for my bad English. 
Thank you!
 A: Note that $A$ is the additive group of vector space of dimension $n$ ove $F_q$ is the same to say that $A$ is an elementary abelian $p$-group of order $p^n$. Hence to prove this problem, you need to prove $A_1=\{a \in A | \chi(a)=1 \}$ is a subgroup of order $p^{n-1}$, i.e. $|A:A_1|=p$. Now applying the hint 1) and 2), you can prove this.
(hint: By 1) $A/A_1$ is faithful and irreducible, hence $A/A_1$ is cyclic. Since $A$ is elementary, we get $|A:A_1|=p$.)
A: See if this makes sense. It took me embarrassingly long to figure out what was going on because I forgot that a linear character $\chi : (A,+) \to (\mathbb{C},\times)$ is a homomorphism from an additive group to a multiplicative group.
It might help to clarify the relationship between $A$ and $V$. One definition I learned for a vector space $V$ over a field $\mathbb{F}$ is that it is an abelian group $(V, +)$ equipped with a scalar multiplication $\cdot_a : \mathbb{F} \to V$ for each element $a \in \mathbb{F}$, such that $\cdot_a$ is a homomorphism from $(\mathbb{F}, \cdot)$ to $(V, +)$. Using this definition, it is clear that the additive group $A$ of $V$ is nothing more than $(V, +)$. Thus, any subgroup $A_1 \leq A$ leads very directly to a subspace $V_1 \subseteq V$. That's why they can ask us to show that $\{ a: \chi(a) = 1 \}$ is a subspace of $V$; elements of $A$ must be vectors. I sincerely hope that this makes statement 2) above fairly obvious; subspaces of $V$ with dimension $n-k$ are in bijection with subgroups of $A$ with index $k$
Now, let $\chi : A \to \mathbb{C}$ be a given, nontrivial, irreducible character of $A$. Since $A$ is abelian, we know that $\chi$ is linear and thus a homomorphism into $(\mathbb{C}, \times)$. In fact, we can say more. It just so happens that the image of $\chi$ is simply the set $\{ e^{2i\pi k/p} : k = 0, 1, \ldots p-1 \}$ of complex $p$-th roots of unity, whose cardinality is $p$ (If this is new to you, I suggest you try to find some characters of $\mathbb{Z}_3^2$).
Let $K = \{ a \in A : \chi(a) = 1 \}$ so that $K = \ker(\chi)$. Since $K$ is the kernal of a homomorphism, it is an additive subgroup of $A$, and thus a subspace of $V$. By the first isomorphism theorem, we see that Im$(\chi) \cong A/K$, where Im($\chi)$ has cardinality $p$. Thus $[A : K] = p$, and you can use part 2) suggested by the book.
