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MIT PRIMES/Art of Problem Solving

CROWDMATH 2024: Generalizations of the Notion of Primes

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Generalizations of the Notion of Primes J
Exercise 0.2
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Thayaden
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Let $\{S_a \}_{a\in n}$ be a submonoid of $M$, where $a$ runs over $n$ is the index set. Let $S$ be the intersection of all possible submonoids of $M$. Notice now we just need to prove $S\subseteq M$. Let $e$ be the identity of $M$ notice that $e \in S_a$ and in turn $e\in S$ therfor $S$ has the identity of $M$ the first characteristic of a submonoid. Now we just need to prove that it is a monoid with respect to the operation of $M$. Let $p,q \in S$ by intersection rules $p\in S_a$ and $q \in S_a$ and notice $S_a$ is a submonoid in itself and thus the operation $p*q\in S_a$ for every $a$ thus $p*q\in S$ by intersection rules again. Thus $S$ is a submonoid in itself!

Let $\{S_a \}_{a\in n}$ be a subgroup of $M$, where $a$ runs over $n$ is the index set. Let $S$ be the intersection of all possible subgroups of $M$. Notice now we just need to prove $S\subseteq M$. Let $e$ be the identity of $M$ notice that $e \in S_a$ and in turn $e\in S$ therfor $S$ has the identity of $M$ the first characteristic of a subgroup. Now we need to prove that it is a subgroup with respect to the operation of $M$. Let $p,q \in S$ by intersection rules $p\in S_a$ and $q \in S_a$ and notice $S_a$ is a subgroup in itself and thus the operation $p*q\in S_a$ for every $a$ thus $p*q\in S$ by intersection rules again. Finally, we need to show that every element has an inverse. Let $x\in S$ and let $x^{-1}$ represent the inverse of $x$. Notice if $x\in S$ then it follows that $x\in S_a$ although notice that all $S_a$ are subgroups of $M$ and thus $x^{-1}\in S_a$ and since $x^{-1}\in S_a$ for any $a\in n$ then by intersection $x^{-1}\in S$ thus $S$ is a subgroup of $M$
This post has been edited 3 times. Last edited by Thayaden, Sep 16, 2024, 6:31 PM
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Exercise 0.2
Thayaden   0
Sep 16, 2024
Let $\{S_a \}_{a\in n}$ be a submonoid of $M$, where $a$ runs over $n$ is the index set. Let $S$ be the intersection of all possible submonoids of $M$. Notice now we just need to prove $S\subseteq M$. Let $e$ be the identity of $M$ notice that $e \in S_a$ and in turn $e\in S$ therfor $S$ has the identity of $M$ the first characteristic of a submonoid. Now we just need to prove that it is a monoid with respect to the operation of $M$. Let $p,q \in S$ by intersection rules $p\in S_a$ and $q \in S_a$ and notice $S_a$ is a submonoid in itself and thus the operation $p*q\in S_a$ for every $a$ thus $p*q\in S$ by intersection rules again. Thus $S$ is a submonoid in itself!

Let $\{S_a \}_{a\in n}$ be a subgroup of $M$, where $a$ runs over $n$ is the index set. Let $S$ be the intersection of all possible subgroups of $M$. Notice now we just need to prove $S\subseteq M$. Let $e$ be the identity of $M$ notice that $e \in S_a$ and in turn $e\in S$ therfor $S$ has the identity of $M$ the first characteristic of a subgroup. Now we need to prove that it is a subgroup with respect to the operation of $M$. Let $p,q \in S$ by intersection rules $p\in S_a$ and $q \in S_a$ and notice $S_a$ is a subgroup in itself and thus the operation $p*q\in S_a$ for every $a$ thus $p*q\in S$ by intersection rules again. Finally, we need to show that every element has an inverse. Let $x\in S$ and let $x^{-1}$ represent the inverse of $x$. Notice if $x\in S$ then it follows that $x\in S_a$ although notice that all $S_a$ are subgroups of $M$ and thus $x^{-1}\in S_a$ and since $x^{-1}\in S_a$ for any $a\in n$ then by intersection $x^{-1}\in S$ thus $S$ is a subgroup of $M$
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Thayaden
Sep 16, 2024
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Exercise 0.2
elee310   0
Jul 28, 2024
What I've tried so far:

$\text{1) A submonoid must (1) include the identity of M and (2) be closed under the binary operation of M}$
$\text{Let A and B be submonoids of M, then } e \in A \land e \in B \rightarrow e \in A \cap B$
$\text{This meets the condition for (1)}$
$\forall f,g \in A \cap B, h=f*g \land i=g*f \in A \text{ by (2), and the same for B}$
$h,i \in A \land h,i \in B \rightarrow h,i \in A \cap B$
$\text{This meets the condition for (2)}$
$\therefore \forall \text{ submonoids } A,B \subseteq M \rightarrow A \cap B \text{ is also a submonoid of } M$

$\text{2) Since subgroups are submonoids, (1) and (2) are given.}$
$\text{Therefore, (3) invertibility of all elements must be satisfied for the intersection to be a subgroup}$
$\text{Let A and B be subgroups of M}$
$\forall f \in A \cap B, g \text{ s.t. } f*g=g*f=e \in A \text{ by (3), and the same for B}$
$g \in A \land g \in B \rightarrow g \in A \cap B$
$\text{This meets the condition for (3)}$
$\therefore \forall \text{ subgroups } A,B \subseteq M \rightarrow A \cap B \text{ is also a subgroup of } M$
0 replies
elee310
Jul 28, 2024
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Exercise 0.2
hkoo   0
May 21, 2024
What I've tried so far:

<Describe what you have tried so far here. That way, we can do a better job helping you!>

(1)&(2)
Submonoid is a subset of monoid which satifies properties of monoid s.t (1) associativity, (2) identity, (3) closure

Let $I$ be a any intersecition of submonoid. because identity is element of submonoid, $I$ has identity. i.e., (2) identity holds in $I$.

Let $a,b,c \in I$. so $a,b,c$ are elements of submonoid s.t $(a*b)*c=a*(b*c)$. therefore (1) associativity holds in $I$.
by similar argument, (3) closure also holds in $I$. Conclusively, $I$ is submonoid.

Subgroup is a subset of group which satifis properties of group s.t (1) associativity, (2) identity, (3) closure, (4) inverse.


For (1),(2),(3), solving problem proceeds in similar way like above.
Let $I$ be a any intersecition of subgroup. if a is element of $I$, then a is element of any subgroup which has $I$ as a subset. By property (4), there exists $b$ s.t $a*b=b*a=e$. so $b$ is also in $I$.
Conclusively, $I$ is subgroup.

Where I'm stuck:

<Describe what's confusing you, or what your question is here!>
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hkoo
May 21, 2024
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No more topics!
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Exercise 0.2
Thayaden   0
Sep 16, 2024
Let $\{S_a \}_{a\in n}$ be a submonoid of $M$, where $a$ runs over $n$ is the index set. Let $S$ be the intersection of all possible submonoids of $M$. Notice now we just need to prove $S\subseteq M$. Let $e$ be the identity of $M$ notice that $e \in S_a$ and in turn $e\in S$ therfor $S$ has the identity of $M$ the first characteristic of a submonoid. Now we just need to prove that it is a monoid with respect to the operation of $M$. Let $p,q \in S$ by intersection rules $p\in S_a$ and $q \in S_a$ and notice $S_a$ is a submonoid in itself and thus the operation $p*q\in S_a$ for every $a$ thus $p*q\in S$ by intersection rules again. Thus $S$ is a submonoid in itself!

Let $\{S_a \}_{a\in n}$ be a subgroup of $M$, where $a$ runs over $n$ is the index set. Let $S$ be the intersection of all possible subgroups of $M$. Notice now we just need to prove $S\subseteq M$. Let $e$ be the identity of $M$ notice that $e \in S_a$ and in turn $e\in S$ therfor $S$ has the identity of $M$ the first characteristic of a subgroup. Now we need to prove that it is a subgroup with respect to the operation of $M$. Let $p,q \in S$ by intersection rules $p\in S_a$ and $q \in S_a$ and notice $S_a$ is a subgroup in itself and thus the operation $p*q\in S_a$ for every $a$ thus $p*q\in S$ by intersection rules again. Finally, we need to show that every element has an inverse. Let $x\in S$ and let $x^{-1}$ represent the inverse of $x$. Notice if $x\in S$ then it follows that $x\in S_a$ although notice that all $S_a$ are subgroups of $M$ and thus $x^{-1}\in S_a$ and since $x^{-1}\in S_a$ for any $a\in n$ then by intersection $x^{-1}\in S$ thus $S$ is a subgroup of $M$
0 replies
Thayaden
Sep 16, 2024
0 replies
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