field theory - Confusion on Artin's Theorem (Linear Independence of Group Homomorphism)

The Artin's Theorem states as follows:

Let $G$ be a group. and let $f_1,\dots, f_n\colon G\to K^{\times}$ be distinct homomorphisms of $G$ into the multiplicative group of a field. Prove that these functions are linearly independent over $K$ Namely, if $a_1,a_2,\dots\in K^{\times}$, then $\forall g\in G$, $$a_1f_1(g)+\cdots+a_nf_n(g)=0$$ implies $a_1=a_2=\cdots=a_n=0$.

Let $G=\mathbb{Z}_5^{\times}$ group of units $U(5)$, and let $K=\mathbb{Z}_5$, hence $K^{\times}=\mathbb{Z}_5 \sim \{0\}$.

Now, consider $f_1(x)=x$ be the identity map and let $f_2(x)$ defined as follows:

$1\mapsto1\\2\mapsto3\\3\mapsto2\\4\mapsto4.$

$f_1$ is the identity mapping basically from $\mathbb{Z}_5^{\times}$ to $\mathbb{Z}_5^{\times}$ hence trivially homomorphism. We show that $f_2$ is also a homomorphism (note that all these calculations are in modulo $5$):

$f_1(2)f_1(2)=3\cdot3=4=f_1(4)=f_1(2\cdot2) \\ f_1(2)f_1(3)=3\cdot2=1=f_1(1)=f_1(2\cdot3) \\ f_1(3)f_1(3)=2\cdot2=4=f_1(4)=f_1(3\cdot3) \\ f_1(2)f_1(4)=3\cdot4=2=f_1(3)=f_1(2\cdot4) \\ f_1(3)f_1(4)=2\cdot4=3=f_1(2)=f_1(3\cdot4) \\ f_1(4)f_1(4)=4\cdot4=1=f_1(1)=f_1(4\cdot4).$

Then the theorem is clearly false since $f_1(2)+f_2(2)=2+3=0$, which is not linearly independent.

This theorem has been proven like decades ago so it must be my reason that is wrong, am I understanding the theorem incorrect or am I missing something here?

Help would be appreciated!

(Found out that this theorem has been proven here)

Answer

For a set of maps, $f_1, f_2, \dots, f_n$ to be linearly independent over $K^\times$, it must be the case that the only solution to
$$\forall g \in G, k_1 f_1(g) + k_2 f_2(g) + \cdots + k_n f_n(g) = 0_K$$
where the $k_i \in K$ for all $i$ and $0_K$ is the additive identity in $K$, is $k_1 = k_2 = \cdots = k_n = 0$.

Note that this does not say that it is enough that $k_1 f_1(g) + k_2 f_2(g) + \cdots + k_n f_n(g) = 0_K$ for one choice of $g \in G$. It must be the case simultaneously for all $g \in G$ (because this is not a statement about the images of elements being independent; it is a statement about functions being independent).

So, for your two functions to be linearly independent, the following set of equations must only be simultaneously satisfied with the $a_i$ are all zero: $$\begin{align*} a_1 f_1(1) + a_2 f_2(1) = a_1 + a_2 = 0 \\ a_1 f_1(2) + a_2 f_2(2) = 2a_1 + 3a_2 = 0 \\ a_1 f_1(3) + a_2 f_2(3) = 3a_1 + 2a_2 = 0 \\ a_1 f_1(4) + a_2 f_2(4) = 4a_1 + 4a_2 = 0 \\ \end{align*}$$
The first and fourth force $a_1 = -a_2$. But using that in the second gives $a_2 = 0$, so that $a_1 = 0$. I.e., the only way all of these e simultaneously satisfied is when the $a_i$ are all zero.

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If you want this to look like linear algebra, then replace $f_i(g)$ with the vector $v_i = (f_i(g): g \in G)$, where we assume $G$ has been well-ordered and the elements of that vector are ordered in the same way. Then we ask that the set $\{v_i : i=1, \dots, n\}$ is linearly independent over $K$. That is, we do not look at the $f_i$ on a single $g \in G$. Instead, we treat each $f_i$ as equivalent to its $G$-indexed sequence of images.

So, for your two $f$s, the vectors whose independence we need to resolve are

$$(1,2,3,4)$$
and
$$(1,3,2,4) \text{.}$$
Then linear independence requires the solution to
$$a_1 (1,2,3,4) + a_2 (1,3,2,4) = (a_1 + a_2, 2 a_1 + 3a_2, 3a_1 + 2a_2, 4a_1 + 4a_2) = (0,0,0,0)$$
is $a_1 = a_2 = 0$. By the same argument as used above the horizontal divider, the only way this equation is satisfied is when $a_1 = a_2 = 0$.