991: For 2 Subgroups with Different Prime Number Orders, if Generators of Subgroups Are Commutative, Product-of-Prime-Numbers-Ordered Cyclic Subgroup Can Be Constructed

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description/proof of that for 2 subgroups with different prime number orders, if generators of subgroups are commutative, product-of-prime-numbers-ordered cyclic subgroup can be constructed

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Topics

About: group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of cyclic group by element.
• The reader admits the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group.
• The reader admits the proposition that any 2 different prime-number-ordered subgroups share only 1.
• The reader admits the proposition that any principal integral domain is a greatest common divisors domain, and for each 2 elements on the principal integral domain, each of the greatest common divisors of the 2 elements is a one by which the sum of the principal ideals by the 2 elements is the principal ideal.

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Target Context

• The reader will have a description and a proof of the proposition that for any group and its any 2 subgroups with any different prime number orders, if any generators of the subgroups are commutative, a the-product-of-the-prime-numbers-ordered cyclic subgroup can be constructed.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$G$: $\in \{\text{ the groups }\}$
$p$: $\in \{\text{ the prime numbers }\}$
$q$: $\in \{\text{ the prime numbers }\}$
$G_p$: $\in \{\text{ the }p\text{ -ordered subgroups of }G\}$, $=\{g_p,...,g_p^p=1\}$
$G_q$: $\in \{\text{ the }q\text{ -ordered subgroups of }G\}$, $=\{g_q,...,g_q^q=1\}$
$G_{pq}$: $=\{g_p{g}_q,...,(g_p{g}_q{)}^{pq}\}$
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Statements:
(
$p\ne q$
$\wedge$
$g_p{g}_q=g_q{g}_p$
)
$⟹$
(
$G_{pq}\in \{\text{ the }(pq)\text{ -orderd cyclic subgroups of }G\}$
$\wedge$
$\mathrm{\forall }j,k\in \mathbb{N}\text{ such that }1\le j<p\text{ and }1\le k<q(\{g_p^j{g}_q^k,...,(g_p^j{g}_q^k{)}^{pq}\}=G_{pq})$
)
//

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2: Note

$G_p$ inevitably takes the shape of $\{g_p,...,g_p^p=1\}$, by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group. It is likewise for $G_q$.

By the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group, $G_p$ or $G_q$ can be regarded to be generated by any $g_p^j$ or $g_q^k$ where $1\le j<p$ and $1\le k<q$, but $g_p^j{g}_q^k=g_q^k{g}_q^j$ holds anyway, because $g_p^j{g}_q^k=g_p^{j-1}{g}_p{g}_q{g}_q^{k-1}=g_p^{j-1}{g}_q{g}_p{g}_q^{k-1}=g_p^{j-2}{g}_p{g}_q{g}_p{g}_q^{k-1}=g_p^{j-2}{g}_q{g}_p{g}_p{g}_q^{k-1}=g_p^{j-2}{g}_q{g}_p^2{g}_q^{k-1}=...=g_q{g}_p^j{g}_q^{k-1}=g_q^2{g}_p^j{g}_q^{k-2}=...=g_q^k{g}_p^j$.

The 2nd half of Statements is saying that we cannot have any different $(pq)$-ordered cyclic subgroup by just choosing a different generator: this proposition says that $g_p^j{g}_q^k$ generates a $(pq)$-ordered cyclic subgroup, but in fact, the subgroup is no different from $G_{pq}$.

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3: Proof

Whole Strategy: Step 1: see that $G_{pq}$ is a $(pq)$-ordered cyclic subgroup; Step 2: see that $\{g_p^j{g}_q^k,...,(g_p^j{g}_q^k{)}^{pq}\}=G_{pq}$.

Step 1:

Let us see that $G_{pq}$ is a $(pq)$-ordered cyclic subgroup.

$1\in G_{pq}$: $(g_p{g}_q{)}^{pq}=g_p^{pq}{g}_q^{pq}=(g_p^p{)}^q(g_q^q{)}^p=1^q{1}^p=1$.

For each $(g_p{g}_q{)}^j,(g_p{g}_q{)}^k\in G_{pq}$, $(g_p{g}_q{)}^j(g_p{g}_q{)}^k=(g_p{g}_q{)}^{j+k}$. $j+k=mpq+l$ where $m,l\in \mathbb{N}$ and $0\le l<pq$. $(g_p{g}_q{)}^{j+k}=(g_p{g}_q{)}^{mpq+l}=(g_p{g}_q{)}^{mpq}(g_p{g}_q{)}^l=((g_p{g}_q{)}^{pq}{)}^m(g_p{g}_q{)}^l=(g_p^{pq}{g}_q^{pq}{)}^m(g_p{g}_q{)}^l=((g_p^p{)}^q(g_q^q{)}^p{)}^m(g_p{g}_q{)}^l=(1^q{1}^p{)}^m(g_p{g}_q{)}^l=1^m(g_p{g}_q{)}^l=(g_p{g}_q{)}^l$. When $1\le l<pq$, $(g_p{g}_q{)}^l\in G_{pq}$ and when $l=0$, $(g_p{g}_q{)}^0=1=(g_p{g}_q{)}^{pq}\in G_{pq}$. So, $(g_p{g}_q{)}^j(g_p{g}_q{)}^k\in G_{pq}$ anyway.

For each $(g_p{g}_q{)}^j\in G_{pq}$, $(g_p{g}_q{)}^{pq-j}\in G_{pq}$, and $(g_p{g}_q{)}^j(g_p{g}_q{)}^{pq-j}=(g_p{g}_q{)}^{pq}=1$ and $(g_p{g}_q{)}^{pq-j}(g_p{g}_q{)}^j=(g_p{g}_q{)}^{pq}=1$.

The associativity holds because it holds in the ambient $G$.

So, $G_{pq}$ is a subgroup of $G$.

Let us see that $\{g_p{g}_q,...,(g_p{g}_q{)}^{pq}\}$ is distinct.

Let us suppose that $(g_p{g}_q{)}^j=(g_p{g}_q{)}^k$ where $1\le j\le k\le pq$ without loss of generality. $1=(g_p{g}_q{)}^{k-j}$, $g_p^{k-j}=g_q^{j-k}$, because $g_p{g}_q=g_q{g}_p$, $=1$ by the proposition that any 2 different prime-number-ordered subgroups share only 1, so, $k-j=mp=nq$, which means that $k-j=0$: $k-j$ is a multiple of $pq$ but $k-j<pq$. So, $j=k$.

So, $G_{pq}$ is a $(pq)$-ordered cyclic subgroup.

Step 2:

Let us see that $\{g_p^j{g}_q^k,...,(g_p^j{g}_q^k{)}^{pq}\}=G_{pq}$.

$\{g_p^j{g}_q^k,...,(g_p^j{g}_q^k{)}^{pq}\}$ is a $(pq)$-ordered cyclic subgroup by Step 1, because also $g_p^j$ generates $G_p$ and also $g_q^k$ generates $G_q$, by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group, and Step 1 applies for $\{g_p^j{g}_q^k,...,(g_p^j{g}_q^k{)}^{pq}\}$ with just $g_p$ and $g_q$ replaced with $g_p^j$ and $g_q^k$ respectively.

Let us see that for each $r,s\in \mathbb{N}$ such that $1\le r\le p$ and $1\le s\le q$, $g_p^r{g}_q^s\in G_{pq}$, because then, as each element of $\{g_p^j{g}_q^k,...,(g_p^j{g}_q^k{)}^{pq}\}$ is of that form, it will be that $\{g_p^j{g}_q^k,...,(g_p^j{g}_q^k{)}^{pq}\}\subseteq G_{pq}$.

There are some $n,m\in \mathbb{Z}$ such that $r-s=np+mq$ by the proposition that any principal integral domain is a greatest common divisors domain, and for each 2 elements on the principal integral domain, each of the greatest common divisors of the 2 elements is a one by which the sum of the principal ideals by the 2 elements is the principal ideal, because $\mathbb{Z}$ is a principal integral domain, $gcd(p,q)=\{1\}$, and $1\mathbb{Z}=p\mathbb{Z}+q\mathbb{Z}$.

Let $l:=r-np=s+mq$. Then, $(g_p{g}_q{)}^l=g_p^l{g}_q^l=g_p^{r-np}{g}_q^{s+mq}=g_p^r{g}_p^{-np}{g}_q^s{g}_q^{mq}=g_p^r(g_p^p{)}^{-n}{g}_q^s(g_q^q{)}^m=g_p^r{1}^{-n}{g}_q^s{1}^m=g_p^r{g}_q^s$.

So, $g_p^r{g}_q^s\in G_{pq}$.

So, $\{g_p^j{g}_q^k,...,(g_p^j{g}_q^k{)}^{pq}\}\subseteq G_{pq}$.

As $|\{g_p^j{g}_q^k,...,(g_p^j{g}_q^k{)}^{pq}\}|=|G_{pq}|$ as finite sets, $\{g_p^j{g}_q^k,...,(g_p^j{g}_q^k{)}^{pq}\}=G_{pq}$.

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References

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