Chapter 1 Group Theory

First, we describe the application of groups in revealing structure in other mathematical objects. Then, we study the relationship between groups, culminating in a classification of groups of order $pq$, where $p$ and $q$ are primes.

1.1 Groups and Actions

1.1.1 Beginnings

Definition 1.1.1.

A group $(G,e,\cdot )$ is a set with an associative binary operation, s.t.

• $\forall a,b,c\in G,a\cdot (b\cdot c)=(a\cdot b)\cdot c$

• $\forall g\in G,\exists g^{-1}$ s.t. $g{g}^{-1}=g^{-1}g=e$

• $\forall g\in G,e\cdot g=g\cdot e=g$

Definition 1.1.2.

A subgroup of group $G$ is a subset that is also a group.

Example 1.1.3.

Consider the group $(\mathbb{Z},0,+)$, and a subgroup . We show that $H=n\mathbb{Z}=\{nm\mid m\in \mathbb{Z}\}$, $n\ge 0$.

If $H=\left\{0\right\}$, then we can write $H=0\mathbb{Z}$. Otherwise, consider the set $X=\{h\in H\mid h>0\}$, which is non-empty by assumption. Pick $n=\min X$ (by well-ordering), and for a given $h\in H$, factor it as $h=nq+r$, . Thus, $r\in H$, by closure, but $r\notin X$, thus $r=0$ and $H\subset n\mathbb{Z}$. Moreover, $n\mathbb{Z}\subset H$, since $n\in H$.

Definition 1.1.4.

A group action is an operation $G\times XtoX$, s.t.

• $\forall h,g\in G,x\in X,h\cdot (g\cdot x)=(hg)\cdot x$

• $\forall x\in X,e\cdot x=x$

We say $G$ acts on $X$ (through an action $\cdot$)

Groups interact with sets through actions. Through actions, we can find equivalences in such sets.

Definition 1.1.5.

If $G$ acts on $X$, then the orbit of $x\in X$ is $⟦x⟧=\{g\cdot x\mid g\in g\}$.

Theorem 1.1.6.

If $G$ acts on $X$, then $x\sim y\iff x=g\cdot y$ is an equivalence relation.

Proof.

We check the properties:

• Reflective: $x\sim x$, since $e\cdot x=x$

• Symmetric: $x\sim y$, then $x=g\cdot y$, and $y=e\cdot y=(g^{-1}g)\cdot y=g^{-1}\cdot (g\cdot y)=g\cdot x$

• Transitive: $x\sim y$ and $y\sim z$, then $y=g\cdot x$, $z=h\cdot y$, and $z=h\cdot (g\cdot x)=(hg)\cdot x$

∎

Thus, we find that the orbits partition the set into equivalence classes. This observation is central towards our results for this section.

An important example is when $G$, or is acting on itself, through the group operation (or as we will later see, with conjugation).

Example 1.1.7.

Let be a subgroup. Then $H$ acts on $G$, $h\cdot g\mapsto hg$. Moreover, $a\sim b\iff a=hb\iff a{b}^{-1}=h\iff a{b}^{-1}\in H$. Oftentimes, we write this as $H\cdot g=\{h\cdot g\mid h\in H\}$. In this case, we also call the orbits cosets.

This idea of equivalence classes also lends itself to quotients.

Definition 1.1.8.

Suppose , $H$ acts on $G$. Then, $G/H=\{⟦g⟧\mid g\in G\}=\{H\cdot g\mid g\in G\}$.

Example 1.1.9.

Consider the groups of integers under addition, $(\mathbb{Z},0,+)$ and $n\mathbb{Z}=\{nm\mid m\in \mathbb{Z}\}$. Then,

Example 1.1.10.

We might ask when a quotient $G/H$ is a group. A sufficient condition is called normality. We say $N⊲G$ a normal subgroup if $gN=Ng$, or that $gn{g}^{-1}\in N$ for all $n\in N,g\in G$. Indeed, if $N⊲G$, then the operation $(gN)(hN)\mapsto (gh)N$ is a well-defined operation:

• If $g_{1}N=g_{2}N$, then $g_1=g_{2}n$, and $g_{1}h=g_{2}nh=g_2(g_2^{-1}m{g}_2)h=m{g}_{2}h\in N(g_{2}h)=(g_{2}h)N$, so $g_{1}hN=g_{2}hN$.

• $(g_N)(g^{-1}N)=(g{g}^{-1})N=N$, so we have inverse and identity elements.

In fact, all orbits are also cosets.

Definition 1.1.11.

Let $G$ act on $X$. Then, for $x\in X$, we write $G_x≔\{g\in G\mid g\cdot x=x\}$, the stabilizer of $x$, and is a subgroup.

Theorem 1.1.12.

If $G$ acts on $X$, then $G/G_x\cong G\cdot x$

Proof.

We consider $f:g{G}_x\mapsto g\cdot x$. Then, it is a well-defined bijective function, because

• For $g{G}_x=h{G}_x$, $g=hs$ for some $s\in G_x$, and $f(g{G}_x)=g\cdot x=h\cdot (s\cdot x)=h\cdot x=f(h{G}_x)$.

• For $y\in G\cdot x$, then $y=g\cdot x$, and $f(g{G}_x)=y$.

• For $f(g{G}_x)=f(h{G}_x)$, then $g\cdot x=h\cdot x$, so that $(h^{-1}g)\cdot x=(h^{-1}h)\cdot x=x$, i.e. $h{G}_x=g{G}_x$.

∎

Corollary 1.1.13.

, then the map ${\varphi }_g:H\to gH$ that sends $h\mapsto gh$ is bijective.

Proof.

The stabilizer $H_g=\{h\in H\mid gh=g\}=\left\{e\right\}$, and so $H\cong H/\left\{e\right\}\cong gH$. ∎

Corollary 1.1.14.

, then $\left|G\right|=\left|H\right|\cdot \left|G/H\right|$

Proof.

We know that $G/H$ are a collection of equally sized orbits that partition the set. Thus, the size of the set is a product of the size of each orbit $\left|H\right|$ with the number of orbits $\left|G/H\right|$. ∎

1.1.2 Permutations

We demonstrate the use of our technology on the permutation group.

Definition 1.1.15.

The symmetric group $S_n=\{f:{\mathbb{Z}}_n\to {\mathbb{Z}}_n\mid f\text{<em class="ltx\_emph ltx\_font\_upright"> bijective </em>}\}$ is a group under function composition.

Definition 1.1.16.

$\sigma \in S_n$ is a cycle over its range $x_0,x_1,\mathrm{\dots },x_m$ where $x_0=x_m$, if

• $\sigma (x_i)=x_{i+1}$, for

• $\sigma (y)=y$ for $y\ne x_i$

We say cycles $\sigma ,\tau$ are disjoint if their range are disjoint, i.e. $\sigma (m)\ne m⟹\tau (m)=m$.

The result for this section is to find a representation for the elements in the permutation group,

Theorem 1.1.17.

Every $\sigma \in S_n$ is a unique product of cycles ${\tau }_1\mathrm{\cdots }{\tau }_m$, up to order of ${\tau }_i$.

Lemma 1.1.18.

If cycles $\tau ,\sigma \in S_n$ are disjoint, they commute.

Proof.

Consider any $m\in {\mathbb{Z}}_n$, and assume WLOG that $\tau (x)\ne x$ (and so, $\sigma (x)=x$). First, $\tau \circ \sigma (x)=\tau (x)$. Moreover, since $\tau$ is injective, then $\tau (\tau (x))\ne \tau (x)$, so $\sigma (\tau (x))=\tau (x)$, and $\sigma \circ \tau (x)=\tau (x)$. ∎

Lemma 1.1.19.

If $\mathbb{Z}$ acts on ${\mathbb{Z}}_n$, then for all $x\in X$, $\mathbb{Z}\cdot x\cong Z_m$ for some $m$.

Proof.

Since is a subgroup, then ${\mathbb{Z}}_x=m\mathbb{Z}$ for some $m$, and $\mathbb{Z}\cdot x\cong \mathbb{Z}/m\mathbb{Z}\cong {\mathbb{Z}}_m$. ∎

Now, we prove the decomposition theorem on permutations:

1.1.17.

We prove existence and uniqueness separately.

• existence: Consider an action $\mathbb{Z}$ on ${\mathbb{Z}}_n$, where $p\cdot q\mapsto {\sigma }^p(q)$. Then, for every orbit $⟦s⟧$, we consider the cycle

  Choosing $s_1,\mathrm{\dots },s_m$ representatives for the orbits, then ${\sigma }_{s_1}\mathrm{\cdots }{\sigma }_{s_m}=\sigma$ is our representation.

• uniqueness: Any decomposition of $\sigma$ into cycles induces a partition into orbits, so every decomposition can be formed in the above process. Moreover, since these are exactly the orbits of $\sigma$, every decomposition is the same.

∎

Corollary 1.1.20.

Every permutation $\sigma \in S_n$ is a product of transpositions (i.e. cycles of order 2).

Proof.

It is sufficient to prove for cycles. Indeed, a cycle $(n_1\mathrm{\dots }{n}_m)=(n_1{n}_2)\mathrm{\cdots }(n_{m-1}{n}_m)$. ∎

1.2 Group Homomorphisms

We turn out attention to studying groups and the relationship between groups (instead of how groups interact with other objects). Our primary tool describing such relationship is the homomorphism.

Homomorphism Properties

Definition 1.2.1.

$f:G\to H$ is a homomorphism if for all $g,h\in G$, $f(gh)=f(g)f(h)$.

Definition 1.2.2.

An isomorphism is a bijective homomorphism.

Lemma 1.2.3.

$f(e_G)=e_H$.

Proof.

$e_H=f{(e_G)}^{-1}f(e_G)=f{(e_G)}^{-1}f(e_G{e}_G)=f(e_G)$ ∎

Lemma 1.2.4.

If $f$ is an isomorphism, then $f^{-1}$ is a homomorphism.

Proof.

Let $a=f(a_0),b=f(b_0)$, so that $f^{-1}(ab)=f^{-1}(f(a_0)f(b_0))=f^{-1}f(a_0{b}_0)=a_0{b}_0=f^{-1}(a)f^{-1}(b)$ ∎

Now, we make the relationship between homomorphisms and actions clear. First, observe that $Hom(G,H)$, the set of homomorphisms between $G$ and $H$, and $Aut(G)$, the set of isomorphisms from $G$ to $G$ are groups under composition.

Example 1.2.5.

We show that (some) actions are equivalently homomorphisms between $G$ to an automorphism group.

If $G$ acts on $H$ with $g\cdot (h_1{h}_2)=(g\cdot h_1)(g\cdot h_2)$, then consider $\mu :G\to Aut(H)$, where $\mu (g):h\to g\cdot h$. Indeed, it a homomorphism, since

• $\mu (g)$ is a homomorphism: $\mu (g)(h_1{h}_2)=g\cdot (h_1{h}_2)=(g\cdot h_1)(g\cdot h_2)=\mu (g)(h_1)\mu (g)(h_2)$

• $\mu (g)$ is an automorphism Since it is surjective ($h=\mu (g)(g^{-1}\cdot h)$) and therefore injective.

• $\mu$ is a homomorphism: $\mu (g_1{g}_2)(h)=(g_1{g}_2)\cdot h=g_1\cdot (g_2\cdot h)=\mu (g_1)\circ \mu (g_2)(h)$.

Conversely, if we have a homomorphism $\mu :G\to Aut(H)$, then we define the action $g\cdot h=\mu (g)(h)$, where indeed,

• $e\cdot h=\mu (e)(h)=id(h)=h$

• $g_1\cdot (g_2\cdot h)=\mu (g_1)\mu (g_2)(h)=\mu (g_1{g}_2)(h)=(g_1{g}_2)\cdot h$

Finally, we mention important subgroups induced by a homomorphism.

Definition 1.2.6.

If $f:G\to H$ is a homomorphism, then the kernal $kerf=\{g\in G\mid f(g)=e\}$, and the image $imf=\{f(g)\mid g\in G\}$ are subgroups of $G$ and $H$ respectively.

Theorem 1.2.7.

If $g:G\to H$ homomorphism, then $kerf⊲G$.

Proof.

For any $k\in kerf$, for any $g\in G$, then $f(gk{g}^{-1})=f(g)f(k)f{(g)}^{-1}=f(g)f{(g)}^{-1}=e$, so $gk{g}^{-1}\in \ker f$. ∎

1.2.1 Isomorphism Theorems

We describe common cases for when groups are isomorphic, as summarized by the three isomorphism theorems.

Theorem 1.2.8.

If $f:G\to H$ is surjective homomorphism, then for $K=\ker f$, $G/K\cong H$.

Proof.

Consider the map $\pi :G/K\to H$ sending $gK\to f(g)$.

• If $g_{1}K=g_{2}K$, then $g_1=g_{2}k$, and $\pi (g_{1}K)=f(g_1)=f(g_{2}k)=f(g_2)f(k)=f(g_2)=\pi (g_{2}K)$

• For $g_{1}K,g_{2}K\in G/K$, then $\pi ((g_{1}K)(g_2)K)=\pi ((g_1{g}_2)K)=f(g_1{g}_2)=f(g_1)f(g_2)=\pi (g_{1}K)\pi (g_{2}K)$

• For any $h\in H$, $h=f(g)$, so $\pi (gK)=f(g)=h$

• If $\pi (g_{1}K)=\pi (g_{2}K)$, then $f(g_1)=f(g_2)$, so $g_1{g}_2^{-1}=k$, and $g_{1}K=g_{2}K$.

∎

Theorem 1.2.9.

If , $N⊲G$, then $H/(H\cap N)=(HN)/N$, where $HN=\{hn\mid h\in H,n\in n\}$.

Proof.

Indeed , with $(h_1{n}_1){(h_2{n}_2)}^{-1}=h_1{n}_1{n}_2^{-1}{h}_2=\stackrel{\text{in }H}{\overbrace{(h_1{h}_2)}}\stackrel{\text{in }N}{\overbrace{\left(h_2^{-1}(n_1{n}_2^{-1})h_2\right)}}$.

Additionally, $N⊲HN$, since for any $h\in HN$, $hn{h}^{-1}\in N$.

Finally, $H\cap N⊲H$, since for any $h\in H,n\in H\cap N$, $hn{h}^{-1}\in N$ (normality of $N$) and $hn{h}^{-1}\in H$ (closure of $H$).

Thus, consider the surjective homomorphism $\pi :H\to (HN)/N$ sending $h\to (he)N=hN$, where $K=ker\pi =H\cap N$. Thus, applying the first isomorphism theorem, we obtain the result. ∎

Theorem 1.2.10.

Let $H,N⊲G$, with . Then, $(G/N)/(H/N)\cong G/H$.

Proof.

Indeed, $H/N⊲G/N$, since $(gN)(hN)(g^{-1}N)=(gh{g}^{-1})N=h_{1}N\in H/N$. Thus, considering the surjective map $\pi :G/N\to H/N$ sending $gN\to gH$, with $K=\ker \pi =\{gK\mid gH=H\}=H/K$. Thus, applying the first isomorphism theorem, we obtain the result. ∎

1.2.2 Generators and Representations

In addition to applying isomorphism theorems, we can study the generators of a group.

Definition 1.2.11.

Let $S\subset G$, then the group generated by $S$, , is the smallest subgroup containing $S$.

For example, considering the dihedral group $D_n$, the generators are $\rho$ and $\tau$ // TODO

Moreover, an important way homomorphisms are applied is to change the study of groups to the study of linear transformations. For example, the sign function. // TODO

1.3 Classification of Groups

Example 1.3.1.

Consider a group $G$ of order $p$, where $p$ is prime. Then, for $e\ne g\in G$, then the subgroup $⟨g⟩$ has order $n\ge 2$ (since $e,g\in ⟨g⟩$). But, we know that since $n\mid p$, then $n=p$, and so $G=⟨g⟩\cong Z_p$.

Above, we classified groups of prime order by studying the subgroups of $G$. To extend this idea, we will develop the Sylow theorems to understand important subgroups of $G$. Then, we apply the isomorphism theorems to understand how these subgroups compose together to build up to $G$.

1.3.1 Sylow Theorems

First, we describe the subgroups that we are looking for:

Definition 1.3.2.

Let $\left|G\right|=p^{n}m$, where $p\nmid m$. Then, a p-subgroup $H$ has order $p^k$ for some $k>0$. Moreover, we say $H$ is a p-Sylow subogroup if $k=n$.

Towards this, we also consider the subgroup,

Definition 1.3.3.

The normalizer $N_G(H)=\{g\in G\mid gH{g}^{-1}=H\}$.

Our main tool towards finding such subgroups is going to be through actions and fixed points (contrast: stabilizers).

Definition 1.3.4.

If $G$ acts on $X$, then $x\in X$ is a fixed-point when $\forall g\in G$, $g\cdot x=x$. We denote $X^G$ the fixed points.

Lemma 1.3.5.

If $G$ acts on $X$, $\left|G\right|=p^k$, then $\left|X^G\right|{\equiv }_p\left|X\right|$.

Proof.

Recall that the set of orbits partition $X$. Focusing on an orbit $G\cdot x$, we know that $p^k=\left|G\right|=\left|G/G_x\right|\left|G\cdot x\right|$. Thus, for $G_x\ne G$ (i.e. $x\notin X^G$), $p\mid |G\cdot x|$. So, collecting the fixed points and non-fixed points,

$$\left|X\right|=\left|X^G\right|+\sum _x(G\cdot x){\equiv }_p\left|X^G\right|$$

∎

Now, we are ready to prove the Sylow theorems.

Theorem 1.3.6.

If $\left|G\right|=p^{n}m$ for $p\nmid m$, then $G$ has a $p$-Sylow subgroup.

Proof.

We prove via induction that for any , that there is a subgroup of order $p^k$.

For $k=1$, consider $X=\{(a_1,\mathrm{\dots },a_p)\in G^p\mid a_1\mathrm{\cdots }{a}_p=e\}$. Then, $|.X|=|.G{|}^{p-1}$, since $a_p={(a_1\mathrm{\cdots }{a}_{p-1})}^{-1}$ has $p-1$ degrees of freedom. Consider $f:X\to X$ sending $(a_1,\mathrm{\dots },a_p)\mapsto (a_p,a_1,\mathrm{\dots },a_{p-1})$, and the action ${\mathbb{Z}}_p$ on $X$, $k\cdot x=f^k(x)$. We find a non-trivial fixed point of $f$ is an element $(a,\mathrm{\dots },a)$, where $\left|⟨a⟩\right|=p$. Indeed, for each of the $N$ non-fixed orbits, each having order exactly $p$, $0{\equiv }_p\left|X\right|{\equiv }_p\left|X^{{\mathbb{Z}}_p}\right|+pN{\equiv }_p\left|X^{{\mathbb{Z}}_p}\right|$. Thus, $\left|X^{{\mathbb{Z}}_p}\right|\ge p$ (non-empty).

Otherwise, for , suppose we have $K$, a subgroup of order $p^{k-1}$. Then, consider the action $K$ on $G/K$, where $p\cdot (gK)\mapsto (ag)K$, and the normalizer $N(K)/K$.

• $N(K)/K\cong {(G/K)}^K$, since $gK\in N(K)/K\iff \forall a\in K,ag=g{k}_a$, i.e. $gK$ is a fixed-point.

• $\left|N(K)/K\right|=\left|{(G/K)}^K\right|{\equiv }_p\left|G/K\right|=p^{n-k+1}m{\equiv }_{p}0$

Thus, choosing a subgroup $⟨bK⟩$ of order $p$ in $N(K)/K$, let $\pi :N(K)\to N(K)/K$, sending $g\mapsto gK$. Constructing $P={\pi }^{-1}(⟨bK⟩)$, $\ker \pi =K$ and so $\left|P\right|=\left|P/K\right|\left|K\right|=p(p^{k-1})=p^k$. ∎

Theorem 1.3.7.

Let . Then, $X=G\cdot P=\{gPg-1\mid g\in G\}$.

Proof.

Let $S,P\in X$. We show that we can pick $a$ s.t. $aS{a}^{-1}\subset P$ (since $\left|S\right|=\left|P\right|$).

We know $S$ acts on $G/P$, with $s\cdot (gP)=(sg)P$. Then, $\left|{(G/P)}^S\right|{\equiv }_p\left|G/P\right|{\not\equiv }_{p}0$, so pick any $a\in {(G/P)}^S$, so that for all $s$, $saP=aP$, i.e. $a^{-1}Sa\subset P$. ∎

Theorem 1.3.8.

Let $n_p$ be the number of p-Sylow subgroups, and $P$ a p-Sylow group. Then, $n_p\mid |G/P|$ and $n_p{\equiv }_{p}1$.

Proof.

Let $X$ be the set of p-Sylow subgroups, with $G$ acting on $X$ by conjugation. Then, $X=G\cdot P$, so $n_p=\left|X\right|=\left|G\cdot P\right|=\left|G\right|/\left|G_P\right|$. Since, , then $p^n\mid G_P$, and so $n_p\mid \left|G\right|/\left|G_P\right|\mid \left|G/P\right|$.

Moreover, acts on $X$, with $\left|X\right|{\equiv }_p\left|X^P\right|$. Charactizing $\left|X^P\right|$, consider any $S\in X^P$, so that $\forall p\in P,pS{p}^{-1}=S$, i.e. . Moreover, since , then $P$ and $S$ are p-Sylow subgroups of $N(S)$, and thus, $P=nS{n}^{-1}$. Finally, since $n\in N(S)$, $P=nS{n}^{-1}=S$, and $1=\left|X^P\right|{\equiv }_p\left|X\right|=n_p$. ∎

1.3.2 Semi-Direct Products

Now that we have developed our key subgroups, we focus on how to compose them.

Definition 1.3.9.

Let $N⊲G$, , where $NH=G$ and $N\cap H=\left\{e\right\}$. Then $N⋊H$ is a group over the set $N\times H$ where $(n_1,h_1)(n_2,h_2)=(n_1(h_1{n}_2{h}_1^{-1}),h_1{h}_2)$.

Recall the similarities between the semi-direct product and the direct-sum from linear algebra. The operation is forced in order for $f:(n,h)\mapsto nh$ to be a homomorphism, since

$$f(n_1,h_1)f(n_2,h_2)=(n_1{h}_1)(n_2{h}_2)=(n_1{h}_1{n}_2{h}_1^{-1})(h_1{h}_2)=f((n_1,h_1)(n_2,h_2))$$

Theorem 1.3.10.

$N⋊H\cong G$ (indeed is a group).

Proof.

The identity element is $(e_N,e_H)$, where $(n,h)(h^{-1}{n}^{-1},h^{-1})=(nh{h}^{-1}{n}^{-1},h{h}^{-1})=e$. Moreover, the operation is associative,

$(n_1,h_1)((n_2,h_2)(n_3,h_3))$$=(n_1,h_1)(n_2(h_2\cdot n_3),h_2{h}_3)$

$=(n_1(h_1\cdot (n_2(h_2\cdot n_3))),h_1{h}_2{h}_3)$

$=(n_1(h_1\cdot n_2)(h_1{h}_2\cdot n_3),h_1{h}_2{h}_3)$

$=(n_1(h_1\cdot n_2),h_1{h}_2)(n_3,h_3)$

$=((n_1,h_1)(n_2,h_2))(n_3,h_3)$

Finally, consider homomorphism $\pi :N⋊H\to G$ sending $(n,h)\mapsto nh$, where

• Injective, since $\ker \pi =\{(n,h)\mid n=h^{-1}\}\subset N\cap H=\left\{e\right\}$

• Surjective, since $NH=G$

and so indeed, isomorphic. ∎

1.3.3 Classification of Groups of order pq

We conclude our chapter on group theory with a quick classification of some groups.

Theorem 1.3.11.

If $\left|G\right|=p^2$, then $G\cong {\mathbb{Z}}_p⋊{\mathbb{Z}}_p$ or $G\cong {\mathbb{Z}}_{p^2}$.

Proof.

If there is some $⟨g⟩=G$, then $G\cong {\mathbb{Z}}_{p^2}$.

Otherwise, consider $G$ acting on itself via conjugation. Choose $e\ne g\in G^G$, and $h\in G\setminus G^G$. Indeed, .

Now, we verify that

• $⟨h⟩\cap ⟨g⟩=\left\{e\right\}$, since $\left|⟨h⟩\cap ⟨g⟩\right|\mid p$ but $⟨h⟩\not\subset ⟨g⟩$.

• $⟨h⟩⟨g⟩=G$, since the map $\pi :(h^k,g^{\mathrm{\ell }})\mapsto h^k{g}^{\mathrm{\ell }}$ is injective.

∎

Theorem 1.3.12.

If $\left|G\right|=pq$, $p>q$, then $G\cong {\mathbb{Z}}_p⋊{\mathbb{Z}}_q$.

Proof.

Consider $P,Q$ subgroups of order $p$ and $q$ respectively. We know that $n_p\mid q$, and since $n_p{\equiv }_{p}1$, then $n_p=1$. Thus,

• $P⊲G$, since $gP{g}^{-1}$ is p-Sylow (conjugation is iso), and so $gP{g}^{-1}=P$

• $P\cap Q=\left\{e\right\}$, since $\left|P\cap Q\right|\le \gcd (p,q)=1$

• $PQ=G$, since $PQ/P\cong Q/(Q\cap P)$ and so $\left|PQ\right|=\left|P\right|\left|Q\right|=\left|G\right|$.

∎

Chapter 2 Ring Theory

2.1 Integers

We outline our path on the integers. First, recall this useful property of the integers.

Theorem 2.1.1.

For every $a,b\in \mathbb{Z}$ where $b\ne 0$, we can factor $a=bq+r$ s.t. .

Proof.

Consider $q=\lfloor \frac{a}{b}\rfloor$. Then, $r=a-bq$ satisifes the inequality. ∎

From the division algorithm, we work towards a decomposition of integers into prime factors.

Definition 2.1.2.

We say $n$ divides $m$, denoted as $n\mid m$ if $m=nq$.

Definition 2.1.3.

The $1\le d=\gcd (a,b)$ if $d\mid a$, $d\mid b$, and moreover for any other $c\mid a$, $c\mid b$ implies $c\mid d$.

For example, for $x\ne 0$, then $\gcd (0,x)=x$. However, note that $\gcd (0,0)$ is not defined, since all integers are common divisors.

We show that the $\gcd (a,b)$ is a well-defined function on $a\ne 0$.

Proof.

Let $S=\{ua+vb\mid ua+vb>0\}$. In particular, $a^2\in S$, so $S$ is non-empty.

Now, considering $d=\min S$, then if $c\mid a$ and $c\mid b$, then $c\mid ua+vb=d$. Moreover, for any other $d^{\prime }$ that claims to be the gcd, then $d^{\prime }\mid d$ and $d\mid d^{\prime }$ shows that $d=d^{\prime }$ (since positive). ∎

Definition 2.1.4.

We say $x$ is a unit if $xy=1$ for some $y$.

Definition 2.1.5.

We say $x$ is an irreducible if $x=ab$ only when $a$ or $b$ is a unit.

Definition 2.1.6.

We say $a$ and $b$ are coprime if $a\mid bx⟹a\mid x$. Moreover, $p$ is prime if $p$ is coprime with every other integer.

Now, we relate these concepts.

Theorem 2.1.7.

$a$ and $b$ are coprime if and only if $\gcd (a,b)=1$.

Proof.

Suppose $\gcd (a,b)=1$, i.e. we can find $ax+by=1$. Then, when $a\mid bc$, then $a\mid (a(xc)+bc(y))=(ax+by)c=c$.

Conversely, if $a$ and $b$ are coprime, then let $d$ be a common divisor, so that $a=xd$ and $b=yd$. Now, since $a\mid ya=xyd=bx$, then $a\mid x$. But also, $x\mid a$, so $x=\pm a$ and $d=\pm 1$. ∎

Theorem 2.1.8.

$p$ is prime if and only if $p$ is irreducible.

Proof.

Suppose $p$ is prime, and write $p=ab$, i.e. $p\mid ab$. WLOG, assume that $p\mid a$, and since $a\mid p$, then $a=\pm p$, i.e. $b=\pm 1$ and $b$ is a unit.

Conversely, suppose $p$ is irreducible and let $p\mid ab$. Then, since the only divisors of $p$ are $\pm 1$ or $\pm p$, then $\gcd (p,a)=1$ or $\gcd (p,a)=p$. If $\gcd (p,a)=1$, then $a$ and $p$ are coprime, and so $p\mid ab⟹p\mid b$. Otherwise, if $\gcd (p,a)=p$, then $a=\pm p$ and $p\mid a$. ∎

Now, we can talk about factorization into irreducibles, i.e. primes.

Definition 2.1.9.

$a$ and $b$ are associates if $a=ub$, for $u$ a unit.

Theorem 2.1.10.

If $0\ne a$ not a unit, then $a=u{p}_1\mathrm{\cdots }{p}_n$, where $u$ is a unit and $p_1,\mathrm{\dots },p_n$ are irreducible. Moreover, if $a=v{q}_1\mathrm{\cdots }{q}_m$, then $n=m$ and we can reorder $q$ with $q_i$ and $p_i$ being associates.

Proof.

We proceed by induction on $|.a|$. If $a$ is an irreducible, then we win. Otherwise, $a$ is reducible, and writing $a=bc$, applying the induction hypothesis proves the existence of factorization.

Arguing uniqueness, we induct on $\max (n,m)$. Since $p_1\mid a$, then $p_1\mid q_i$, say $q_1$. Then, $p_1$ and $q_1$ are associates (since both irreducible), so inducting on $a^{\prime }=v^{\prime }{q}_2\mathrm{\cdots }{q}_m=u^{\prime }{p}_2\mathrm{\cdots }{p}_m$, we win. ∎

2.2 Domains

Definition 2.2.1.

A (commutative) ring $R$ is an abelian group $(R,+,0)$ with an operation $R\times R\to R$,

$$ab=baa(b+c)=ab+ac1r=r$$

Definition 2.2.2.

$0\ne r\in R$ is a zero-divisor if $\exists s\ne 0$ s.t. $sr=0$.

Definition 2.2.3.

We say $R$ is a domain if there are no zero-divisors.

Note that a field is a domain, since units can not be zero-divisors. Moreover, finite domains are fields, since for $a\ne 0$, we have the bijection $f_a:x\mapsto ax$, with $a\cdot f_a^{-1}(1)=1$.

Here, we only concern ourselves with domains, since whenever $ab=ac$ for $a\ne 0$, then $a(b-c)=0=b-c$, i.e. $b=c$.

Definition 2.2.4.

We say $R$ is an Euclidean domain with a function $\delta :R\setminus \left\{0\right\}\to \mathbb{N}$ s.t. for any $a,b\in R$ with $b\ne 0$

For convenience, we denote $\delta (0)=-\mathrm{\infty }$.

Definition 2.2.5.

We say $R$ is a unique-factorization domain if $\exists !r=u{p}_1\mathrm{\cdots }{p}_n$, up to associates.

The main result is to show that in general, Euclidean domains are UFDs. Towards this, we introduce an intermediate concept of ideals.

Definition 2.2.6.

An ideal $I⊲R$ is a subgroup where $ri\in I$ for all $r\in R$, $i\in I$.

Definition 2.2.7.

$I$ is principal if $I=⟨c⟩$ (as a group). $R$ is a principal ideal domain if all ideals are principal.

Considering the subgroup $R/I$, it is moreover a ring with the operation $(r+I)(s+I)=rs+I$. We also define the following operations on ideals (generating new ideals):

Definition 2.2.8.

$$I_1+I_2=\{i_1+i_2\mid i_j\in I_j\}{I}_1{I}_2=\{i_1{i}_2\mid i_j\in I_j\}$$

Lemma 2.2.9.

$⟨c⟩=Rc=\{rc\mid r\in R\}$.

Proof.

We know $rc\in ⟨c⟩$, by definition, so $Rc\subset ⟨c⟩$. Moreover, $Rc$ is an ideal, since $rc-sc=(r-s)c\in Rc$ and $s(rc)=(sr)c\in Rc$. ∎

Note that the integers are Euclidean domains $\delta (r)=|.r|$, as well as PIDs (since only subgroups are $n\mathbb{Z}$), and UFDs.

Theorem 2.2.10.

Euclidean Domains are PIDs.

Proof.

Let $I$ be an ideal. If $I=0$, then $I=⟨0⟩$ is principal.

Otherwise, let $S=\{\delta (i)\mid 0\ne i\in I\}$, non-empty, and pick $d=\mathrm{argmin}S$. Writing $i=dq+r$, we know $r=i-dq\in I$. But, since , then $\delta (r)\notin S$ and so $r=0$. Thus, $S\subset ⟨d⟩$, and so $S=⟨d⟩$. ∎

To show that PIDs are UFDs, we need to prove relationships between primes and irreducibles.

Lemma 2.2.11.

$p$ is prime if and only if $R/⟨p⟩$ is a domain.

Proof.

Note that $p\mid ab\iff (a+⟨p⟩)(b+⟨p⟩)=0+⟨p$. Thus, $p$ prime if and only if one of the terms is zero, the defining property of a domain. ∎

Lemma 2.2.12.

In $R$ PID, and $c$ is irreducible, then $R/⟨c⟩$ is a field (and domain).

Proof.

If $⟨c⟩=R$, then $R/⟨c⟩=0$, a field.

Otherwise, consider any non-zero element $a+⟨c⟩$, i.e. $a\notin ⟨c⟩$. Now, let $⟨a⟩+⟨c⟩=⟨b⟩$, since we are in a PID. Since $0=0a\in ⟨a⟩$, then $⟨c⟩\subset ⟨b⟩$, so write $c=bx$. Moreover, $x$ is not a unit, since $a\in ⟨b⟩\setminus ⟨c⟩$ (i.e. ideals are not the same), so $b$ is a unit. Thus, $⟨b⟩=R$, with $r=(r{b}^{-1})b\in ⟨b⟩$. ∎

Theorem 2.2.13.

In a PID, then irreducibles are equivalent to primes.

Proof.

If $c$ irreducible, then $R/⟨c⟩$ is a domain, so $c$ prime.

If $c$ prime, and if $c=ab$, then suppose $c\mid a$. Then, $a=cq$, $c=cqb$ and $qb=1$, i.e. $b$ is a unit. ∎

Theorem 2.2.14.

PIDs are UFDs.

Proof.

First, we show existence of a factorization, via the algorithm

1. If $r$ irreducible, stop.

2. Otherwise, $r=ab$, both not units, and recurse.

For any path $a_1\mathrm{\dots }$ we get an ascending chain of ideals $⟨a_1⟩\subset \mathrm{\dots }$. Considering $J=\cap ⟨a_i⟩$ is an ideal, then $J=⟨a_n⟩$, i.e. the stopping point. Thus, the algorithm terminates.

Arguing uniqueness, we induct on $\max (n,m)$. Since $p_1\mid a$, then $p_1\mid q_i$, say $q_1$. Then, $p_1$ and $q_1$ are associates (since both irreducible), so inducting on $a^{\prime }=v^{\prime }{q}_2\mathrm{\cdots }{q}_m=u^{\prime }{p}_2\mathrm{\cdots }{p}_m$, we win. ∎

We conclude this section with two examples of Euclidean domains, which we proved are also UFDs.

Example 2.2.15.

Let $\mathbb{Z}[i]=\{a+bi\mid a,b\in \mathbb{Z}\}$, and define $\delta (z)=|.z{|}^2$. Note that $\delta (zw)=\delta (z)\delta (w)$. For any $a,b\in \mathbb{Z}[i]$, with $b\ne 0$, we need to show that $a=bq+r$, with .

For any $z\in ℂ$, we can find $q\in \mathbb{Z}[i]$ where , by choosing the closest lattice point. Thus, choosing $q\in \mathbb{Z}[i]$ the closest point to $a/b$, and $r=a-qb$, we win.

Example 2.2.16.

Consider $F[x]$, the (finite degree) polynomials over a field $F$, and the function $\delta (f)=\mathrm{deg}f$ (with $\delta (0)=-\mathrm{\infty }$). Note that $\delta (fg)=\delta (f)+\delta (g)$. For any $f,g\in F[x]$ with $m=\delta g\ge 0$, we need to show $f=gq+r$ with .

Inducting on $n=\delta (f)$, if , then $f=g0+r$ wins. Otherwise, write $f^{\prime }=f-(f_n{g}_m^{-1})x^{n-m}g$, and inducting, $f^{\prime }=g{q}^{\prime }+r$. Now, $f=g(q^{\prime }+f_n{g}_m^{-1}{x}^{n-m})+r$ wins.

2.3 Chinese Remainder Theorem

First, we prove the first isomorphism theorem for rings.

Definition 2.3.1.

$f:R\to S$ is a ring homomorphism if it is a group homomorphism with $f(rs)=f(r)f(s)$

Theorem 2.3.2.

If $f:R\to S$ is a surjective ring homomorphism, then $R/\ker f\cong S$

Proof.

We know that they are isomorphic as groups, with $\pi :(r+K)\mapsto f(r)$ the group isomorphism. Moreover, it is also a ring isomorphism: $\pi ((r+\ker f)(s+\ker f))=\pi (rs+\ker f)=rs=\pi (r+\ker f)\pi (s+\ker f)$. ∎

Now, we are ready for the Chinese Remainder Theorem.

Theorem 2.3.3.

Let $I_1,\mathrm{\dots },I_k$ be ideals s.t. $I_i+I_j=R$ for all pairs $1\le i,j\le k$. Then,

$$R/(I_1\mathrm{\cdots }{I}_k)\cong (R/I_1)\times \mathrm{\cdots }\times (R/I_k)$$

Lemma 2.3.4.

If $I_1+I_2=R$, then $I_1{I}_2=I_1\cap I_2$.

Proof.

We know $I_1{I}_2\subset I_1\cap I_2$. Moreover, letting $i_1+i_2=1$, then for any $i\in I_1\cap I_2$, then $i=i(i_1+i{i}_2)\in I_1{I}_2$. ∎

CRT.

If $k=1$, then the statement is trivial. For $k=2$, consider the homomorphism

$$f:R\to R/I_1\times R/I_{2}r\mapsto (r+I_1,r+I_2)$$

Let $i_1+i_2=1$, for $i_j\in I_j$. For any $(y_1+I_1,y_2+I_2)$, consider $x=i_1{y}_2+i_2{y}_1$, so

$f(x)$$=((i_1{y}_2+i_2{y}_1)+I_1,(i_1{y}_2+i_2{y}_1)+I_2)$

$=(i_1{y}_2+(1-i_1)y_1+I_1,(1-i_2)y_2+i_2{y}_1+I_2)$

$=(y_1+I_1,y_2+I_2)$

and $f$ is surjective. Moreover, $\ker f=\{r\mid r+I_1=I_1,r+I_2=I_2\}=I_1\cap I_2=I_1{I}_2$, so we win.

Finally, for $k>2$, let $J=I_1\mathrm{\cdots }{I}_{k-1}$. Consider $a_j+b_j=1$, for $a_j\in I_j$ and $b_j\in I_k$, for . Then, we can invoke the induction hypothesis, since $1=\prod (a_j+b_j)\in J+I_k$, so $J+I_k=R$. Now,

$$R/(I_1\mathrm{\cdots }{I}_k)=R/J{I}_k\cong R/J\times R/I_k\cong (R/I_1\times \mathrm{\cdots }R/I_{k-1})\times R/I_k$$

∎

We study the implications on PIDs.

Lemma 2.3.5.

For $a,b$ distinct (not associates) irreducibles in $R$ a PID, then $⟨a^n⟩+⟨b^m⟩=R$, for $n,m\ge 0$.

Proof.

Let $⟨a^n⟩+⟨b^m⟩=⟨c⟩$. Then, $a^n\in ⟨c⟩$ so $a^n=cx$. Similarly, $b^m=cy$.

By UFD properties, then $c=u{a}^i=v{b}^j$, for $i\le n$ and $b\le j$, $u$ and $v$ units. But, since $a$ and $b$ are not associates, then $i=j=0$, and $c$ is a unit and $⟨c⟩=R$. ∎

For example,

$$Z_{12}=\mathbb{Z}/⟨12⟩=\mathbb{Z}/(⟨3⟩⟨4⟩)=\mathbb{Z}/⟨3⟩\times \mathbb{Z}/⟨4⟩=Z_3\times Z_4$$

so that given ${⟦a⟧}_3$ and ${⟦a⟧}_4$, we can find ${⟦a⟧}_{12}$ uniquely. In general, we can factor $a$ into distinct irreducibles.

Chapter 3 Module Theory

3.1 Modules

Definition 3.1.1.

A $R$-module $M$ is an abelian group $(M,0,+)$ with an operation $R\times M\to M$ where

$$(r_1+r_2)m=r_{1}m+r_{2}mr(m_1+m_2)=r{m}_1+r{m}_2{r}_1(r_{2}m)=(r_1{r}_2)m{1}_{R}m=m$$

Definition 3.1.2.

$N⊲M$ is a submodule if $rn\in N$.

We can think of modules as groups that are acted on by rings. Or, we can think of modules as vector spaces on rings.

Example 3.1.3.

$R$ is an $R$-module, with the normal multiplication. Submodules are ideals.

Definition 3.1.4.

$f:M\to N$ is a module homomorphism if it is a group homorphism where $f(rm)=rf(m)$.

Definition 3.1.5.

$M/N$ quotient group is a quotient module with operation $r(m+N)=rm+N$.

Theorem 3.1.6.

$f:M\to N$ surjective, then $M/\ker f\cong N$.

Proof.

They are isomorphic as groups, with $\pi (m+K)=f(m)$. Moreover, it is a module homomorphism,

$$\pi (r(m+K))=\pi (rm+K)=f(rm)=rf(m)=r\pi (m+K)$$

∎

Moreover, we have all our familiar linear algebra concepts.

Definition 3.1.7.

$S\subset M$, then the span of $S$, $⟨S⟩=\{\sum r_i{s}_i\mid r_i\in R,s_i\in S\}$

Definition 3.1.8.

$S$ generates $M$ if $⟨S⟩=M$.

Definition 3.1.9.

$S$ is linearly independent if $\sum r_i{s}_i=0⟹r_i=0$.

Definition 3.1.10.

$S$ is a basis if it is a linearly independent generator.

With this, we note some basic properties of modules.

Lemma 3.1.11.

$M\cong R^n/N$, for some $N⊲M$.

Proof.

Let $M$ be generated by $\{x_1,\mathrm{\dots },x_m\}$. Consider $f:R^n\to M$ by $(r_1,\mathrm{\dots },r_m)\to \sum r_i{x}_i$, surjective homomorphism. Then, $\ker f$ is an submodule, and so we win. ∎

In particular, if $M$ is cyclic, i.e. generated by one element, then $M\cong R/I$ and we can think of it as a ring.

Finally, we reintroduce the direct sum.

Definition 3.1.12.

Then, $N_1\oplus \mathrm{\dots }\oplus N_k$ is a module over the group $N_1\times \mathrm{\dots }\times N_k$, $r(n_1,\mathrm{\dots },n_k)\mapsto (r{n}_1,\mathrm{\dots },r{n}_k)$

Lemma 3.1.13.

Let $N_1,\mathrm{\dots },N_k⊲M$, where $⟨{\cup }_{1\le i\le k}{N}_i⟩=M$, and $N_j\cap (⟨{\cup }_{i\ne j}{N}_i⟩)=0$. Then, $N_1\oplus \mathrm{\dots }{N}_k\cong M$.

Proof.

Consider the map $(n_1,\mathrm{\dots },n_k)\mapsto \sum n_i$. Then, it is surjective, since $⟨{\cup }_{1\le i\le k}{N}_i⟩=M$. Moreover, it is injective, since if $\sum n_i=0$, then $n_i={\sum }_{i\ne j}-n_j\in N_i\cap (⟨{\cup }_{i\ne j}{N}_j⟩)=0$. ∎

3.2 Bases and Dimension

Modules without bases are easy to construct. For example, consider ${\mathbb{Z}}_n$ as a $\mathbb{Z}$-module, where since $nx=0$, then there are no linearly independent sets.

Definition 3.2.1.

$M$ is free if it has a basis.

Definition 3.2.2.

$m\in M$ is torsion if $rm=0$ for $r\ne 0$. $M^{\text{tors}}=\{m\mid m\text{ torsion}\}$, and say $M$ is torsion if $M=M^{\text{tors}}$.

Definition 3.2.3.

If $M$ is free, then $dimM$ is the size of a basis.

Lemma 3.2.4.

In a domain, $M^{\text{tors}}⊲M$

Proof.

It is a subgroup, since for $m,n$ torsion with $rm=sn=0$, then $rs(m-n)=0$ where $rs\ne 0$. Moreover, for any $x\in R$, $r(xm)=x(rm)=0$, so $xm$ also torison. ∎

If $M^{\text{tors}}\ne 0$, then $M$ is not free. Similarly, if $M$ is free, then $M^{\text{tors}}=0$. Later, we will show that every module $M$ is uniquely decomposed into a torsion and free part, but now we focus on studying the free component, in particular proving that dimension is well-defined.

We reduce the question to vector spaces.

Definition 3.2.5.

$I$ ideal, then $IM=\{im\mid i\in I,m\in M\}⊲M$.

Definition 3.2.6.

$M/IM$ is an $R/I$ module, with $(r+I)(m+IM)\mapsto rm+IM$.

Theorem 3.2.7.

$M$ free, then $M/IM$ is free, with same sized bases.

Proof.

Let $S=\{x_1,\mathrm{\dots },x_m\}$ be a basis of $M$, and consider $S^{\prime }=\left\{x_i+IM\right\}$. First, note that $IM=⟨\{i_j{x}_j\mid i_j\in I\}$.

Now, for any $m+IM\in M/IM$, writing $m=\sum r_j{x}_j$, then $m+IM=\sum (r_j+I)(x_j+IM)$, so $S^{\prime }$ is generating. Moreover, if $\sum (r_j+I)(x_j+IM)=im+IM=0$, where $im=\sum i_j{x}_j$, then $\sum (r_j-i_j)x_j=0$, and because $x_j$ are independent, $r_j+I=I$. ∎

In a PID, we can choose an irreducible $c$ and let $I=⟨c⟩$, so that $R/I$ is a field. In general, we can always choose an ideal (say, “maximal”) where $R/I$ is a field, but proof omitted here (requires axiom of choice).

Theorem 3.2.8.

$M$ free $R$-module, then $dimM$ is well-defined.

Proof.

Consider any two bases $\left\{x_i\right\}$ and $\left\{y_i\right\}$ of $M$. Then, $\left\{x_i+IM\right\}$ and $\left\{y_i+IM\right\}$ are bases of $M/IM$, which is a vector space. But, in a vector space, $dimM/IM$ is well-defined, so the size of the two bases are the same. ∎

Proving that dimension is well-defined in a vector space is easy by considering the change of basis matrix. It is clearly invertible, and thus applying Gaussian elimination, its reduced row-echelon form is $I$, i.e. is square.

Theorem 3.2.9.

In a PID, $M$ free, then $N⊲M$ is also free, with $n=dimN\le dimM=m$. Moreover, there exists a basis $\{a_1{x}_1,\mathrm{\dots },a_n{x}_n\}$ and $\{x_1,\mathrm{\dots },x_m\}$ of $N$ and $M$ respectively.

Proof.

It is sufficient to prove for $M=R^m$. Since $N⊲M$, then $N=⟨(a_1,\mathrm{\dots },a_n)⟩$, since $R^m$ is a PID. Thus, $\left\{a_i{e}_i\right\}$ and $\left\{e_i\right\}$ are our basis for $N$ and $M$ (ignoring $a_i=0$). ∎

This is a weaker than incomplete basis for vector spaces, which states that any $N$-basis $N$ extends to a $M$-basis.

3.3 Structure Theorem

Theorem 3.3.1.

In a PID, $\exists !T,F⊲M$ s.t. $M=T\oplus F$, where $T$ torsion and $F$ free.

Proof.

We prove existence and uniqueness separately.

Existence: Let $T=M^{\text{tors}}$, and $F^{\prime }=M/T$, with $T$ torsion. Then, $F_{}^{\prime }{}_{}{}^{\text{tors}}=0$, since for any $m+T$ torsion, then $r(m+T)=rm+T=0$, so $rm$ torsion, and so $m$ torsion and $m+T=0+T$. Now, write $F^{\prime }\cong ⨁R/⟨a_i⟩=F$. With $\left\{(0,\mathrm{\dots },1+⟨a_i⟩,\mathrm{\dots },0)\right\}$, generating $F$, and since $a_i(1+⟨a_i⟩)=0$, but $F^{\prime }$ torsion-free, then $a_i=0$ and $F$ free.

Thus, $T$ are the slots of $R^m$ where $a_i\ne 0$ and $F$ are the slots where $a_i=0$, and $T\oplus F=M$.

Uniqueness: Suppose $M=T_1\oplus F_1=T_2\oplus F_2$. Then, $T_1\subset M^{\text{tors}}$. Moreover, for any $m=t_1+f_1$ torsion with $rm=0$, $r\ne 0$, then $r{t}_1=-r{f}_1$, so $r{f}_1=0$ and $f_1=0$. Thus, $T_1=M^{\text{tors}}$, and similarly $T_2=M^{\text{tors}}=T_1$.

Now, consider ${\varphi }_{F_i}:M\to F_i$ surjective homomorphism, sending $t+f_i\mapsto f_i$. Then, ${\varphi }_{F_1}={\varphi }_{F_2}=\text{id}-{\varphi }_T$, so $F_1={\varphi }_{F_1}(M)={\varphi }_{F_2}(M)=F_2$. ∎

Note that in general, $M/N\oplus N\cong ̸M$, for example $4\mathbb{Z}\cong ̸4\mathbb{Z}/2\mathbb{Z}\oplus 2\mathbb{Z}\cong 2\mathbb{Z}\oplus 2\mathbb{Z}$, since only the former is cyclic.

We can refine this theorem further.

Theorem 3.3.2.

In a PID, $\exists !⟨p_1⟩,\mathrm{\dots },⟨p_m⟩$ (not necessarily distinct) irreducibles and $e_1,\mathrm{\dots },e_m>0$ where

$$M\cong R^n\oplus R/⟨p_1^{e_1}⟩\mathrm{\cdots }\oplus R/⟨p_m^{e_m}⟩$$

Proof.

The free part is fixed. Moreover, the torsion part is isomorphic to some $R/⟨a_1⟩\oplus \mathrm{\cdots }\oplus R/⟨a_n⟩$. Now, we may write each $⟨a_1⟩=⟨p_1^{e_{11}}\mathrm{\cdots }{p}_n^{e_{1n}}⟩$, and applying the Chinese remainder theorem, we get existence.

Arguing uniqueness, we know the $p_i$ irreducibles are unique, since these are the terms in the ring for which there are zero vector-divisors (i.e. $pm=0$ for some $m\in M$). Moreover, if ${⨁}_{i}R/⟨p_i^{e_i}⟩\cong {⨁}_{i}R/⟨p_i^{f_i}⟩$, and ordering $e_i$ and $f_i$ in ascending order, let $\varphi$ be the isomorphism. Considering ${\varphi }^{\prime }=\varphi |R/⟨p_1^{e_1}⟩$, then $\text{im }{\varphi }^{\prime }=R/⟨p_1^{f_1}⟩$, so $f_1=e_1$ and we win via induction. ∎

3.3.1 Classification of Finite Abelian Groups

Theorem 3.3.3.

Let $A$ be a finite abelian group, and $\prod p_i^{n_i}$ the prime factorization of $\left|A\right|$. Then, for some $\sum e_{ij}=n_j$,

$$A\cong \prod \prod \mathbb{Z}/⟨p_j^{e_{ij}}⟩$$

Proof.

Consider $A$ as a $\mathbb{Z}$-module, with $n\cdot g=g^n$. Then, we win via the structure theorem. ∎

For example, there are six abelian groups of order $200$, since $200=2^3{5}^2$.

3.3.2 Jordan Normal Form

Theorem 3.3.4.

Let $V$ be a $ℂ$-module and $T:V\to V$ a linear transformation. Then, $V$ is a direct sum of generalized eigenspaces for $T$, i.e. $T({\beta }_i)={\lambda }_i{\beta }_i+{\beta }_{i+1}$ or $T({\beta }_i)={\lambda }_i{\beta }_i$.

Proof.

Consider $V$ as a $C[x]$-module, where $f\cdot v=f(T)(v)$, i.e. $T(f\cdot v)=(xf)(T)(v)$. Write $V{\cong }_{\varphi }⨁ℂ[x]/⟨p_i^{e_i}⟩$.

Because $ℂ$ is algebraically closed, $p_i=x-{\lambda }_i$. Considering the module $ℂ[x]/⟨{(x-\lambda )}^n$, we have a basis (for ), $\left\{{\beta }_i^{\prime }={(x-{\lambda }_i)}^j+⟨{(x-{\lambda }_i)}^n⟩\right\}$. Consider the basis $\left\{{\beta }_i={\varphi }^{-1}({\beta }_i^{\prime })\right\}$ of $V$. Then, $\beta$ is our generalized eigenbasis, since

$T({\beta }_i)$$=x\cdot {\beta }_i={\varphi }^{-1}\varphi (x{\beta }_i)={\varphi }^{-1}\varphi (x\cdot {(x-\lambda )}^i+⟨{(x-\lambda )}^n⟩$

$={\varphi }^{-1}\left((x-\lambda ){(x-\lambda )}^i+\lambda {(x-\lambda )}^i+⟨{(x-\lambda )}^n⟩\right)$

$={\varphi }^{-1}\left({(x-\lambda )}^{i+1}+\lambda {(x-\lambda )}^i+⟨{(x-\lambda )}^n⟩\right)$

∎