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January 22nd

Today I learned some representation theory, from Artin 10.1.1: all representations of finite groups with dimension $1$ have image that is finite cyclic. Unpacking the definitions, we are provided a homomorphism $\rho$ taking\[\rho:G\to\op{GL}_1(\FF)\]for some finite group $G$ and field $\FF.$ We need to show that $\rho(G)$ is finite cyclic.

The main step of the proof is to realize that $\op{GL}_1(\FF)\cong\FF^\times.$ For this we claim that\[(T:\FF\to\FF)\stackrel\varphi\longmapsto T(1)\]is an isomorphism. We run through the conditions.

  • Note $\varphi$ is well-defined because $T(1)\ne0,$ for this would imply $T(x)=xT(1)=0$ for any $x\in\FF,$ so $T$ would not be injective (or surjective).

  • We know $\varphi$ is homomorphic by direct computation: $(T\circ T')(1)=T(T'(1))=T'(1)\cdot T(1).$

  • Continuing, $\varphi$ is injective because $T(1)=T'(1)$ implies $T(x)=xT(1)=xT'(1)=T'(x)$ for any $x\in\FF,$ so $T=T'.$

  • We see $\varphi$ is surjective because, for any $c\in\FF,$ $x\mapsto cx$ is in $\op{GL}_1(\FF)$ (this can be checked manually). This automorphism satisfies $1\mapsto c.$

It follows that $\varphi$ is an isomorphism, so we're done here.

From the above we see that we are really given a homomorphism $\rho:G\mapsto\FF^\times.$ Note that $\rho(G)$ is a finite subgroup of $\FF^\times,$ so we're really showing that finite subgroups of $\FF^\times$ are finite cyclic. With this transformed problem, we rename $\rho(G)$ to $G$ in order to be able to ignore $\rho.$ That is, $G\subseteq\FF^\times$ is a finite subgroup, and we want to show that $G$ is finite cyclic.

There are a couple of ways to finish from here. The cheapest is that Artin always assumes that our base field is $\FF=\CC,$ in which case any finite subgroup of $\CC^\times$ is finite cyclic by roots-of-unity type arguments. For example, take $n=|G|,$ and we can say that\[G\subseteq\left\{x\in\FF^\times:x^n-1=0\right\}\]by Lagrange's theorem on groups. In fact, this is an equality because the right-hand side has at most $n$ elements by Lagrange's theorem on polynomials. But in $\CC^\times,$ we know that the right-hand side is\[\left\langle e^{2\pi i/n}\right\rangle.\]This can be checked by, say, noting that both sets have size $n$ and satisfy $x^n-1=0.$

However, this can be done in significantly more generality, for what we want is true in any field. I might have seen this proof before, but I have since forgotten it. Anyways, we mimic the proof that finite fields have a multiplicative generator and borrow the finish from this post . The main lemma we need is that for elements $g,h\in G,$ there exists an element whose multiplicative order which is the least common multiple of the orders of $g$ and $h.$

This is somewhat obnoxious to show. We begin by claiming that if $a:=\ord(g)$ and $b:=\ord(h)$ are coprime, then $\ord(gh)=ab.$ Indeed, certainly $(gh)^{ab}=1.$ And in the other direction, if $(gh)^x=1,$ then\[1=(gh)^{ax}=h^{ax},\]so $b\mid ax,$ and $b\mid x.$ Similarly, $a\mid x,$ so $ab\mid x.$ It follows $ab$ really is our order. Now, to finish our lemma, we remove the condition that $a$ and $b$ are coprime. However, we remark that we can separate the primes in $a$ and $b$ with unique prime factorization by writing\[a_*=\prod_{\nu_p(a)\ge\nu_p(b)}p^{\nu_p(a)}\qquad\text{and}\qquad b_*=\prod_{\nu_p(b) \gt \nu_p(a)}p^{\nu_p(b)}.\]Namely, $a_*b_*=\lcm(a,b)$ while having $\gcd(a_*,b_*)=1$ because no prime factors are shared. So we only need to remark that $g^{a/a_*}$ will have order $a_*$ and $h^{b/b_*}$ will have order $b_*$ in order to assert an element of order $a_*b_*,$ which is what we wanted.

To finish, we note that, because $G$ is finite, we may inductively say that there exists an element $g$ whose order $m$ is divisible by the order of all elements of $G.$ On one hand, $m\le|G|$ because $m$ is the order of some element of $G.$ On the other hand, every element of $G$ is the root of the polynomial\[x^m-1=0,\]so Lagrange's theorem on polynomials says that $|G|\le m.$ It follows that the order of $g$ is the full $m=|G|,$ so $G$ is finite cyclic. We remark here that this proof is really just the proof that every prime has a primitive root (via field theory), generalized.