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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)

Samantha Y.
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  • Make clear $K,+$ and $K^*,\times$ are two different groups, your $f_1=f_2^{-1}$ so $f_1+f_2 \ne 0$ (as $f_1(1)+f_2(1) = 2 \ne 0$) – reuns May 05 '19 at 01:06
  • I think I still don't get it, let me rephrase my question. – Samantha Y. May 05 '19 at 01:07
  • $K$ should be "$\cong \mathbb{Z}_5$" and $K^\times$ should be "$\cong \mathbb{Z}_5^\times$", not "$= \mathbb{Z}$" and "$= \mathbb{Z}_5$". ("\cong" and "^\times") – Eric Towers May 05 '19 at 01:22

1 Answers1

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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.

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 $a_1 = a_2 = 0$.

Eric Towers
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  • Perhaps this is a totally dumb question ... which one of my map $f_1$, $f_2$ is not a homomorphism? – Samantha Y. May 05 '19 at 01:15
  • @SamanthaY. : Currently, $f_1$ and $f_2$ aren't even maps. – Eric Towers May 05 '19 at 01:24
  • Why are they not maps, I'm mapping them from basically $(\mathbb{Z}_5,\times)\mapsto(\mathbb{Z}_5,+,\times)$. – Samantha Y. May 05 '19 at 01:27
  • @SamanthaY. : Since you say $K = \mathbb{Z}$, the field of units of $K$ is ${-1,1}$, but you have image elements not in this set. This $K$ is also not a field. – Eric Towers May 05 '19 at 01:28
  • The theorem states that the linear independent holds for each $g\in G$, so if I can find one $g\in G$ that it doesn't hold, the theorem falls apart. – Samantha Y. May 05 '19 at 01:28
  • That was a typo, corrected. – Samantha Y. May 05 '19 at 01:28
  • @SamanthaY. : False. This theorem states that a set of functions are linearly independent, not a set of values -- no set of two or more values can be linearly independent over a field. – Eric Towers May 05 '19 at 01:29
  • @SamanthaY. : The use of "$\forall g \in G$" in your quoted block means "the following equality holds simultaneously for every $g \in G$". It is not enough that it hold for one $g \in G$. – Eric Towers May 05 '19 at 01:32
  • @SamanthaY. : Perhaps another way to see it is : "The list of left-hand sides of the following equation over every possible specialization to a $g \in G$ is not just a list of zeroes unless the $a_i$ are all zeroes." – Eric Towers May 05 '19 at 01:34
  • Okay I feel like an idiot, there's something that you're seeing but I'm not seeing ... =( All I can see from this theorem is that given any elements $a_1,a_2,\dots$ from the field, and pick any distinct homomorphisms, then for every $g\in G$, every linear combination of these homormophisms under one $g$ at a time is linearly independent. So $a_1f_1(g)+a_2f_2(g)+\cdots=0\implies a_1=a_2=\cdots=0$. But in above I have $1\cdot f_1(2)+1\cdot f_2(3)=0\not\implies a_1=a_2=0$. – Samantha Y. May 05 '19 at 01:34
  • @SamanthaY. Maybe it's clearer if you write$\ a_1f_1(g)+\cdots+a_nf_n(g)\equiv0$ – saulspatz May 05 '19 at 01:38
  • @SamanthaY. : The values of the $f_i$ are fixed. You are solving for the set of possible $(a_i: i=1\cdots n)$ that satisfy all the equations simultaneously. I've written more about this in both halves of my answer. – Eric Towers May 05 '19 at 01:43
  • Oh lol I got it now, seriously stupid me. I really appreciate your patience in explaining this to me! Thank you for your time. I must look like a big idiot haha – Samantha Y. May 05 '19 at 01:45
  • @SamanthaY. : You're welcome. Have a good weekend! – Eric Towers May 05 '19 at 01:45
  • @EricTowers sorry can't upvote, don't have enough rep yet lol – Samantha Y. May 05 '19 at 01:49
  • @SamanthaY. : No worries. I'm not here for the rep. – Eric Towers May 05 '19 at 01:49