Note - Structure of Finite Commutative Groups

notegroup-theory

Isomorphism Classes of Finite Commutative Groups

Theorem

Let $(G,\cdot )$ be a commutative group with $n$ elements, then there exist unique integers $1<d_m\mid d_{m-1}\mid d_{m-2}\mid \dots \mid d_1$ such that $n=d_1\cdot d_m$ and $G\simeq {\mathbb{Z}}_{d_1}\times {\mathbb{Z}}_{d_2}\times \dots \times {\mathbb{Z}}_{d_m}$

Example 1

List the isomorphism classes of finite commutative groups with 6 elements.

$6=d_1\cdot \dots \cdot d_m$ $d_m\ge 2$ $d_m\mid d_{m-1},d_{m-1}\mid d_{m-2},\dots ,d_2\mid d_1$

$6=6=2\cdot 3$ but $3\nmid 2$.

Therefore by the theorem above, every finite commutative group with 6 elements is cyclic of order $6$ (isomorphic to ${\mathbb{Z}}_6$)

Question

How about ${\mathbb{Z}}_2\times {\mathbb{Z}}_3$?

Answer: We found last time that ${\mathbb{Z}}_2\times {\mathbb{Z}}_3\simeq {\mathbb{Z}}_6$, because $\gcd (2,3)=1$ and recall the Chinese Remainder Theorem: if $n=ab$, $\gcd (a,b)=1$, then ${\mathbb{Z}}_a\times {\mathbb{Z}}_b\simeq {\mathbb{Z}}_n$.

Example 2

List isomorphism classes of commutative groups with 8 elements.

We want to decompose $8=d_1\cdot \dots \cdot d_m$ with $d_m\ge 2$, $d_m\mid d_{m-1},\dots ,d_2\mid d_1$, and in particular $d_1\ge d_2\ge \dots \ge d_m\ge 2$.

The only ways to write 8 as products of integers in weakly decreasing order is,

1. $8=8$
2. $8=4\cdot 2$
3. $8=2\cdot 2\cdot 2$

Thus, by the theorem, if $(G,\cdot )$ is commutative with 8 elements, $G\simeq {\mathbb{Z}}_8$, or ${\mathbb{Z}}_4\times {\mathbb{Z}}_2$, or ${\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2$

Note that ${\mathbb{Z}}_2\times {\mathbb{Z}}_4\simeq {\mathbb{Z}}_4\times {\mathbb{Z}}_2$.

Example 3

$(U(16),\cdot )=\{i\in {\mathbb{Z}}_{16}:\gcd (i,1)=1\}$ (this is equivalent to $i$ odd) $=\{1,3,5,7,9,11,13,15\}$ (note this has 8 elements)

So this is isomorphic to one of those we listed previously, in particular it is isomorphic to ${\mathbb{Z}}_4\times {\mathbb{Z}}_2$.

Why? If not, then it is isomorphic to ${\mathbb{Z}}_8$ or ${\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2$.

If the latter is true, then $i^2\equiv 1$ for all $i\in U(16)$ but $3^2=9\not\equiv 1\mathrm{mod}16$, we have a contradiction.

Otherwise, we check $i^4=1$ for all $i$, so there is no element of order 8, so $U(16)$ is not cyclic. $1^4=1\equiv 1\mathrm{mod}16$ $(-1{)}^4=(-1{)}^4\cdot 1^4=1\cdot 1^4=1$. If $i^4.\equiv 1\mathrm{mod}16$ then $(16-i{)}^4=(-i{)}^4\equiv 1\mathrm{mod}16$ $⟹15^4=(-1{)}^4\equiv 1\mathrm{mod}16$.

$3^4 = 81 =80 + 1 = 16*5 +1 \equiv 1 \bmod 16$
$\implies 1 \equiv (-3)^4=13^4$

$5^4 = 25^2 \equiv 9^2 = 81 \equiv 1 \bmod 16$
$\implies 1 \equiv (-5)^4 \equiv 11^4 \bmod 16$

$9^2=81\equiv 1 \bmod 16 \implies 9^4 \equiv 1^2 \equiv 1 \bmod 16$
$\implies 1 \equiv (-9)^4\equiv 7^4$

Example 4

List isomorphism classes of commutative groups with 120 elements.

Solution: $120=d_1,d_2,\dots ,d_m$ with $d_1\ge d_2\ge \dots \ge d_m\ge 2$ and $d_m\mid d_{m-1}$, $d_{m-1}\mid d_{m-2}$, etc.

$120=12\cdot 10=2^2\cdot 3\cdot 2\cdot 5=2^3\cdot 3\cdot 5$

The divisibility conditions force $2\cdot 3\cdot 5=30\mid d_1$ Options are: $120=120$ $120=60\cdot 2$ $120=30\cdot 2\cdot 2$

Then every commutative group with 120 elements is isomorphic to exactly one of ${\mathbb{Z}}_{120}$, ${\mathbb{Z}}_{60}\times {\mathbb{Z}}_2$, or ${\mathbb{Z}}_{30}\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2$.

Example 5

${\mathbb{Z}}_{12}\times {\mathbb{Z}}_{10}$ is isomorphic to exactly one on the above list.

$12=4\cdot 3⟹{\mathbb{Z}}_{12}\simeq {\mathbb{Z}}_4\times {\mathbb{Z}}_3$ $10=5\cdot 2⟹{\mathbb{Z}}_{10}\simeq {\mathbb{Z}}_5\times {\mathbb{Z}}_2$

${\mathbb{Z}}_{12}\times {\mathbb{Z}}_{10}\simeq {\mathbb{Z}}_4\times {\mathbb{Z}}_3\times {\mathbb{Z}}_5\times {\mathbb{Z}}_2$ (note that 4,3,5 are pairwise coprime)

By the Chinese Remainder Theorem, ${\mathbb{Z}}_4\times {\mathbb{Z}}_3\times {\mathbb{Z}}_5\simeq {\mathbb{Z}}_{60}$

Thus, we have that ${\mathbb{Z}}_{12}\times {\mathbb{Z}}_{10}\simeq {\mathbb{Z}}_{60}\times {\mathbb{Z}}_2$.

Proof of Uniqueness

If $G\simeq {\mathbb{Z}}_{d_1}\times \dots \times {\mathbb{Z}}_{d_m}$ and $G\simeq {\mathbb{Z}}_{f_1}\times \dots \times {\mathbb{Z}}_{f_m}$

We have, $d_1\ge d_2\ge \dots \ge d_m\ge 2$ with $m\ge 1$ $f_1\ge f_2\ge \dots \ge f_r\ge 2$ with $r\ge 1$

$d_m\mid d_{m-1}$, $d_{m-1}\mid d_{m-2}$, etc. $f_r\mid f_{r-1}$, $f_{r-1}\mid f_{r-2}$, etc.

We want that $r=m$ and $d_i=f_i$ for all $i$.

Observation: $d_1=$ smallest $\{u\ge 1:g^u=e_G\forall g\in G\}$ $⋮$

Example 6

We saw that we have ${\mathbb{Z}}_{120}$, ${\mathbb{Z}}_{60}\times {\mathbb{Z}}_2$, and ${\mathbb{Z}}_{30}\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2$ are the only isomorphism classes of commutative groups with 120 elements. These all have different exponents (120, 60, and 30, respectively) which means they are not isomorphic.

Example 7

Every finite commutative group with 16 elements is isomorphic to either ${\mathbb{Z}}_{16}$, ${\mathbb{Z}}_8\times {\mathbb{Z}}_2$, ${\mathbb{Z}}_4\times {\mathbb{Z}}_4$, ${\mathbb{Z}}_4\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2$, or ${\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2$.

The exponents of each of these are respectively, 16, 8, 4, 4, and 2. Note that ${\mathbb{Z}}_4\times {\mathbb{Z}}_4$ and ${\mathbb{Z}}_4\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2$ have the same exponent.

How many elements of order 4 do they each have?

In ${\mathbb{Z}}_4\times {\mathbb{Z}}_4$, $(1,i),(3,i),(i,1),(i,3)$ have order 4 for any $i\in {\mathbb{Z}}_4$. Excluding repetitions, this leaves us with 12 such elements. In ${\mathbb{Z}}_4\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2$ $(1,i,j),(3,i,j)$ have order 4 for any $i,j\in {\mathbb{Z}}_2$. This leaves us with 8 such elements.

Thus, we have that ${\mathbb{Z}}_4\times {\mathbb{Z}}_4$ and ${\mathbb{Z}}_4\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2$ are not isomorphic.

Proof of Existence

$G\simeq {\mathbb{Z}}_{d_1}\times \dots \times {\mathbb{Z}}_{d_m}$

First Step: We break $G$ into simpler parts.

$n=p_1^{{\alpha }_1}\cdot \dots \cdot p_r^{{\alpha }_r}$ where $p_1,\dots p_r$ are distinct primes, ${\alpha }_i\ge 1$ for all $i$.

Using the fact from (P-Torsion Subgroup) that $g,g^{\prime }\in G[p]⟹{\mathrm{ord}}_{}(g)=p^k$, ${\mathrm{ord}}_{}(g^{\prime })=p^{k^{\prime }}$ for some $k,k^{\prime }\in \mathbb{N}⟹{\mathrm{ord}}_{}(g{g}^{\prime })\mid \mathrm{lcm}(p^k,p^{k^{\prime }})=\max (k,k^{\prime })⟹{\mathrm{ord}}_{}(g{g}^{\prime })=p^{k^{\prime \prime }}$ for some $k^{\prime \prime }\le \max (k,k^{\prime })⟹G[p]$ is stable under products.

(b) Assume $p\mid |G|\stackrel{\text{Cauchy}}{⟹}\exists g\in G,{\mathrm{ord}}_{}(g)=p=p^1⟹e\ne g\in G[p]⟹G[p]\ne \{e\}$.

Conversely, if $G[p]\ne \{e\}⟹\exists e\ne g\in G[p]⟹{\mathrm{ord}}_{}(g)=p^k$ for some $k>0\stackrel{\text{Lagrange}}{⟹}{p}^k\mid |G|⟹p\mid |G|.$

Next time we show $G\simeq G[p_1]\times \dots \times G[p_r]$.

Show $G\simeq G[p_1]\times \dots \times G[p_r]$

Define $\phi :G[p_1]\times \dots \times G[p_r]\to G$ $p_1,\dots ,p_r$ are the distinct prime factors of $n=|G|$. $\phi (a_1,\dots ,a_r)=a_1\bullet \dots \bullet a_r$

Because $G$ is commutative, we have easily that $G$ is a morphism.

Assume $(a_1,\dots ,a_r)\in \ker \phi$. $⟹a_1\cdot \dots \cdot a_r=e_G⟹a_1=a_2^{-1}\cdot a_3^{-1}\cdot \dots \cdot a_r^{-1}$.

Let ${\mathrm{ord}}_{}(a_i)=p_i^{k_i}$ for $k_i\ge 0$ $⟹{\mathrm{ord}}_{}((a_i^{-1}))={\mathrm{ord}}_{}(a_i)=p_i^{k_i}$

$p_1^{k_1}={\mathrm{ord}}_{}(a_2^{-1}\cdot \dots \cdot a_r^{k_r})=$ divisor of $\mathrm{lcm}({\mathrm{ord}}_{}(a_2^{-1}),{\mathrm{ord}}_{}(a_r^{-1}))$ since $a_i$ commute, this is a divisor of $p_2^{k_2},p_r^{k_r}$ $⟹p_1^{k_1}=1⟹k_1=0$, so $a_1=e_G$.

An inductive argument implies that $a_i=e_G$ for all $i$, which implies $\ker \phi =\{e_G,\dots ,e_G\}$, thus $\phi$ is injective.

Now show $\phi$ is surjective. Let $g\in G$ where $|G|=n$. By Lagrange’s theorem, we have that $g^n=e_G$

$n=p_1^{{\alpha }_1}\cdot \dots \cdot p_r^{{\alpha }_r}$ which are $r$ pairwise coprime factors.

By the Chinese Remainder Theorem, ${\mathbb{Z}}_n\underset{\underset{\psi \text{ isomorphism}}{\leftarrow }}{\simeq }{\mathbb{Z}}_{p_1}^{{\alpha }_1}\times \dots \times {\mathbb{Z}}_{p_r}^{{\alpha }_r}$

\begin{align*} p_{1}&= \psi(1,0,\ldots,0)\\ &\vdots \\ p_{r} &= \psi(0,0,\ldots,0,1) \end{align*} $$ Note, the $p_{i}$ above could be a $\beta_{i}$... (Missed some lines that reach the below) In $\mathbb{Z}\implies \gamma_{1}\beta_{1} + \ldots + \gamma_{r}\beta_{r}=1 \bmod n \implies \gamma_{1}\beta_{1} + \ldots +\gamma_{r}\beta_{r}=1+nk$ for some $k\in \mathbb{Z}$. $g=g^{-1}$ and $g^n = e$ $\implies g=g^{1+nk} = g^{\gamma_{1}\beta_{1}+\ldots + \gamma_{r} \beta_{r}}= g^{\gamma_{1}\beta_{1}}\cdot\ldots \cdot g^{\gamma_{r}\beta_{r}}$ *Question:* What is the order of $g^{\beta_{1}}$? $(1,0,\ldots,0) \in \mathbb{Z}_{p_{2}}\alpha_{1}\times \ldots \times \mathbb{Z}_{p_{r}}\alpha_{r}$ has order $p_{1}^{\alpha_{1}} \implies \beta_{1}\in \mathbb{Z}_{n}$ also has order $p_{1}^{\alpha_{1}} \bmod n$. $\implies p_{1}^{\alpha_{1}} \cdot \beta_{1} \equiv 0 \bmod n$. $\beta_{1}=n\cdot s$ for some $s\in \mathbb{Z}$ $\implies (g^{{\beta_{1}}})^{p_{1}^{\alpha_{1}}}$ From remark in [[Proof of P-group Theorem]], $G \simeq G[p_{1}] \times\ldots \times G[p_{r}]$ We can use the Chinese Remainder Theorem to regroup as $\mathbb{Z}_{d_{1}}\times \ldots \times \mathbb{Z}_{d_{m}}$, with $d_{m}\mid d_{m-1}\mid\ldots\mid d_{2}\mid d_{1}$ and $d_{1}\cdot \ldots \cdot d_{m} = p$. #### Example Say $G=G[2] \times G[3] \times G[5]$ which respectively are $\mathbb{Z}_{4}\times \mathbb{Z}_{2}, \mathbb{Z}_{9}\times \mathbb{Z}_{9} \times \mathbb{Z}_{3},\mathbb{Z}_{5}\times \mathbb{Z}_{5}$. Arrange the indices as In columns $4,9,5$ are pairwise coprime, as are $2,9,5$, and $1,3,1$. Thus, by the Chinese Remainder Theorem $\mathbb{Z}_{4}\times \mathbb{Z}_{9}\times \mathbb{Z}_{5} \simeq \mathbb{Z}_{4\cdot{9}\cdot{5}} = \mathbb{Z}_{180}$ $\mathbb{Z}_{}$