## 305 -9. Groups

We want to define the notion of group that will be fundamental to determine which polynomials are solvable by radicals. This notion is very important and appears in every area of mathematics.

We motivate the definition through the example that most concerns us: automorphisms of fields. They are particular class of isomorphisms, so we begin with them.

Some of the arguments below have been discussed in previous lectures.

1. Isomorphisms

Recall that a field homomorphism is a map ${h:{\mathbb F}\rightarrow{\mathbb I}}$ between fields ${{\mathbb F}}$ and ${{\mathbb I}}$ such that ${h(a+b)=h(a)+h(b)}$ and ${h(ab)=h(a)h(b)}$ for all ${a,b\in{\mathbb F}.}$ This implies that ${h(0)=0}$ and that ${h(-a)=-h(a)}$ for all ${a\in{\mathbb F}.}$

There are two possibilities for the value of ${h(1),}$ because ${h(1)=h(1^2)=h(1)^2,}$ so either ${h(1)=0}$ or else ${h(1)=1.}$ In the first case, ${h(a)=h(a1)=h(a)h(1)=h(a)0=0}$ for all ${a\in{\mathbb F}.}$ We will assume from now on that ${h(1)=1}$ to avoid this degenerate case. Note that this implies that ${h(a^{-1})=h(a)^{-1}}$ for all nonzero ${a\in{\mathbb F}.}$

The following observation is obvious:

Lemma 1 Suppose ${{\mathbb F}:{\mathbb K}}$ and ${{\mathbb H}:{\mathbb I}:{\mathbb J}.}$ Suppose that ${h:{\mathbb F}\rightarrow{\mathbb I}}$ is a field homomorphism.

Then ${h:{\mathbb F}\rightarrow{\mathbb H}}$ is a field homomorphism.

If ${j=h\upharpoonright{\mathbb K}}$ is the restriction of ${h}$ to ${{\mathbb K},}$ then ${j:{\mathbb K}\rightarrow{\mathbb I}}$ is a field homomorphism.

Let ${h''{\mathbb F}=\{h(a):a\in{\mathbb F}\}.}$ If ${h''{\mathbb F}\subseteq{\mathbb J},}$ then ${h:{\mathbb F}\rightarrow{\mathbb J}}$ is a field homomorphism. ${\Box}$

We are interested in which sets can be the images of homomorphisms.

Lemma 2 Suppose that ${h:{\mathbb F}\rightarrow{\mathbb I}}$ is a field homomorphism. Then ${h''{\mathbb F}}$ is a field.

Proof: We need to check that ${h''{\mathbb F}}$ is a subfield of ${{\mathbb I}.}$ First note that it is nonempty, and it is closed under addition and multiplication. This is because ${0=h(0)\in h''{\mathbb F}.}$ Also, if ${\alpha,\beta\in h''{\mathbb F},}$ then there are ${a,b\in{\mathbb F}}$ such that ${\alpha=h(a)}$ and ${\beta=h(b)}$ and therefore ${\alpha+\beta=h(a)+h(b)=h(a+b)\in{\mathbb F}.}$ Similarly, ${\alpha\beta=h(ab)\in{\mathbb F}.}$

One also checks easily that ${h''{\mathbb F}}$ is closed under additive inverses and under multiplicative inverses of nonzero elements. $\Box$

A field homomorphism ${h:{\mathbb F}\rightarrow{\mathbb I}}$ is an isomorphism iff ${h}$ is a bijection.

Lemma 3 If ${h:{\mathbb F}\rightarrow{\mathbb I}}$ is an injective homomorphism, then ${h:{\mathbb F}\rightarrow h''{\mathbb F}}$ is an isomorphism. ${\Box}$

We are mostly concerned with isomorphisms between subfields of ${{\mathbb C}.}$ The following observation shows that any isomorphism always has many fixed points:

Lemma 4 Suppose ${h:{\mathbb F}\rightarrow{\mathbb I}}$ is a homomorphism and that ${{\mathbb F}}$ and ${{\mathbb I}}$ are subfields of ${{\mathbb C}.}$ Let

$\displaystyle {\rm Fix}(h)=\{a\in{\mathbb F}:h(a)=a\}.$

Then ${{\rm Fix}(h)}$ is a field.

Proof: ${0\in{\rm Fix}(h).}$ By our convention, ${1\in{\rm Fix}(h).}$ Note that ${{\rm Fix}(h)}$ is closed under addition and multiplication, under additive inverses, and under multiplicative inverses of nonzero elements. For example, if ${a\in{\rm Fix}(h)}$ and ${a\ne0,}$ then ${h(a^{-1})=h(a)^{-1}=a^{-1},}$ so ${a^{-1}\in{\rm Fix}(h).}$ $\Box$

Since every subfield of ${{\mathbb C}}$ contains ${{\mathbb Q},}$ it follows that every homomorphism between subfields of ${{\mathbb C}}$ fixes ${{\mathbb Q}.}$

We are interested in the case where ${[{\mathbb F}:{\mathbb Q}]<\infty.}$

In what follows, even if I forget to state this explicitly, all fields we consider are subfields of ${{\mathbb C}}$ unless stated otherwise.

Lemma 5 Suppose that ${{\mathbb F}={\mathbb K}(\alpha)}$ and that ${\alpha}$ is algebraic over ${{\mathbb K}.}$ Suppose that ${h:{\mathbb F}\rightarrow{\mathbb I}}$ is an injective homomorphism, and that ${{\mathbb K}\subseteq{\rm Fix}(h).}$ Let ${\beta=h(\alpha).}$ Then

$\displaystyle h:{\mathbb K}(\alpha)\rightarrow{\mathbb K}(\beta)$

is an isomorphism, and it is given as follows: Any element of ${{\mathbb K}(\alpha)}$ can be written in a unique way in the form

$\displaystyle a=b_0+b_1\alpha+\dots+b_{n-1}\alpha^{n-1},$

where ${b_0,\dots,b_{n-1}\in{\mathbb K}}$ and ${n={\rm deg}(p_\alpha)}$ where ${p_\alpha\in{\mathbb K}[x]}$ is the minimal polynomial of ${\alpha}$ over ${{\mathbb K}.}$ Then

$\displaystyle h(a)=b_0+b_1\beta+\dots+b_{n-1}\beta^{n-1}.$

Proof: Note first that since ${{\mathbb K}\subseteq{\rm Fix}(h)\subseteq{\mathbb I}}$ and ${\beta=h(\alpha)\in h''{\mathbb F}\subseteq{\mathbb I},}$ then ${{\mathbb K}(\beta)\subseteq{\mathbb I}.}$

Given ${a\in{\mathbb F}={\mathbb K}(\alpha),}$ we can write ${a}$ as stated above. Then

$\displaystyle \begin{array}{rcl}h(a)&=&h(b_0)+h(b_1\alpha)+\dots+h(b_{n-1}\alpha^{n-1})\\ &=&h(b_0)+h(b_1)h(\alpha)+\dots+h(b_{n-1})h(\alpha^{n-1})\\ &=&h(b_0)+h(b_1)h(\alpha)+\dots+h(b_{n-1})h(\alpha)^{n-1}\\ &=&b_0+b_1\beta+\dots+b_{n-1}\beta^{n-1},\end{array}$

and this number clearly belongs to ${{\mathbb K}(\beta).}$

This shows that ${h:{\mathbb K}(\alpha)\rightarrow{\mathbb K}(\beta).}$ Note also that if

$\displaystyle p_\alpha(x)=a_0+a_1x+\dots+a_{n-1}x^{n-1}+a_nx^n,$

where ${a_0,\dots,a_n\in{\mathbb K},}$ then ${0=p_\alpha(\alpha)}$ so

$\displaystyle 0=h(p_\alpha(\alpha))=a_0+a_1\beta+\dots+a_n\beta^n=p_\alpha(\beta).$

This means that ${\beta}$ is a root of ${p_\alpha.}$ Since ${p_\alpha}$ is irreducible over ${{\mathbb K}[x],}$ then ${p_\alpha=p_\beta.}$ It follows that ${[{\mathbb K}(\beta):{\mathbb K}]=n.}$

From this we deduce that any element ${\gamma}$ of ${{\mathbb K}(\beta)}$ can be written in the form

$\displaystyle b_0+b_1\beta+\dots+b_{n-1}\beta^{n-1}$

for some ${b_0,\dots,b_{n-1}\in{\mathbb K},}$ and therefore ${h}$ is onto. It is also 1-1, by assumption. Therefore, ${h}$ is a bijection. $\Box$

This shows that any isomorphic image of ${{\mathbb Q}(\alpha)}$ must have the form ${{\mathbb Q}(\beta)}$ for some ${\beta}$ that is also a root of ${p_\alpha.}$ This also means that there are at most ${n}$ injective homomorphisms with domain ${{\mathbb Q}(\alpha),}$ where ${n={\rm deg}(p_\alpha).}$ By results from lecture 7.3, if ${\beta}$ is any root of ${p_\alpha,}$ the map ${h:{\mathbb Q}(\alpha)\rightarrow{\mathbb Q}(\beta)}$ defined as above is an isomorphism. We conclude that there are exactly ${n}$ injective homomorphisms with domain ${{\mathbb Q}(\alpha),}$ thanks to the following theorem:

Theorem 6 If ${p\in{\mathbb K}[x]}$ is irreducible, then all the roots of ${p}$ are simple. In particular, all the roots of the minimal polynomial ${p_\alpha}$ of any ${\alpha}$ algebraic over ${{\mathbb K}}$ are distinct.

Proof: Suppose that ${t}$ is a double root of ${p,}$ and that ${p}$ is irreducible. Then, over ${{\mathbb C},}$ ${p(x)=(x-t)^2q(x)}$ for some polynomial ${q\in{\mathbb C}[x].}$ Note that since ${t}$ is not necessarily in ${{\mathbb K},}$ then ${q}$ is not necessarily in ${{\mathbb K}[x].}$

It follows that the derivative ${p'(x)=2(x-t)q(x)+(x-t)^2q'(x),}$ so ${t}$ is also a root of ${p'.}$ Note also that ${p'\in{\mathbb K}[x].}$ Since ${p}$ and ${p'}$ have a common root, they cannot be relatively prime, because then there would be polynomials ${r,s}$ such that ${1=pr+p's,}$ but this equation gives a conradiction when evaluated at ${t.}$

It follows that there is a nonconstant polynomial ${h\in{\mathbb K}[x]}$ that divides both ${p}$ and ${p'.}$ But then ${{\rm deg}(h)\le{\rm deg}(p')<{\rm deg}(p),}$ and this contradicts that ${p}$ is irreducible. $\Box$

For example, if ${h}$ is an injective homomorphism with domain ${{\mathbb Q}(\sqrt{5})}$, then ${h(\sqrt{5})=\sqrt{5}}$ or ${h(\sqrt{5})=-\sqrt{5}.}$ In the first case, ${h}$ is the identity. In the second, ${h}$ is the map

$\displaystyle h(a+b\sqrt{5})=a-b\sqrt{5}$

for any rational numbers ${a,b.}$

As another example, the only fields isomorphic to ${{\mathbb Q}({\root3\of7})}$ are ${{\mathbb Q}({\root3\of7}),}$ ${{\mathbb Q}({\root3\of7}\zeta_3),}$ and ${{\mathbb Q}({\root3\of7}\zeta_3^2).}$

The following follows from the same arguments as above:

Theorem 7 Suppose that ${{\mathbb F}:{\mathbb K}}$ and that ${\alpha\in{\mathbb F}}$ is algebraic over ${{\mathbb K}}$ with minimal polynomial ${p_\alpha.}$ Let ${h}$ be any injective homomorphism with domain ${{\mathbb F}}$ with ${{\mathbb K}\subseteq{\rm Fix}(h).}$ Then ${h(\alpha)}$ is also a root of ${p_\alpha}$ and ${p_{h(\alpha)}=p_\alpha.}$ ${\Box}$

2. Automorphisms

Definition 8 If ${h:{\mathbb F}\rightarrow{\mathbb F}}$ is an isomorphism, then we say that ${h}$ is an automorphsim of ${{\mathbb F}.}$ The collection of all automorphsism of ${{\mathbb F}}$ is denoted by ${{\rm Aut}({\mathbb F}).}$

For example, ${{\rm Aut}({\mathbb Q})=\{id\},}$ where ${id}$ denotes the identity map.

As another example, ${{\rm Aut}({\mathbb Q}({\root3\of7}))=\{id\}}$ as well.

On the other hand, ${{\rm Aut}({\mathbb Q}(\sqrt{2}))}$ consists of precisely two elements: The identity, and the only homomorphism that sends ${\sqrt2}$ to ${-\sqrt2.}$

In the next lecture we will study automorphisms in some detail, and introduce the notion of group.

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