Why Does Every Element in U(\mathbb{I}_m) Have an Inverse?

[Math Amateur]

I am reading Joseph J. Rotman's book, Advanced Modern Algebra and am currently focused on Chapter 1: Groups I.

I need some help with the proof of Proposition 1.52.

Proposition 1.52 reads as follows:View attachment 4520

I have several related questions that need clarification ...Question 1

In the above text Rotman writes the following:

" ... ... If \(\displaystyle (a,m) = 1\), then \(\displaystyle [a][x] = 1\) can be solved for \(\displaystyle [x]\) in \(\displaystyle \mathbb{I}_m\). ... ... "


Note: Rotman uses \(\displaystyle \mathbb{I}_m\) for the group of integers mod \(\displaystyle m \) and uses \(\displaystyle (a,m)\) for the gcd of \(\displaystyle a\) and \(\displaystyle m\) ...

Can someone explain to me exactly why the above statement by Rotman follows ...Question 2

In the above text Rotman writes the following:

" ... ... Now \(\displaystyle (x,m) = 1\), for \(\displaystyle rx + sm = 1\) for some integer \(\displaystyle s\) and so \(\displaystyle (x,m) = 1\) ... ... "

I cannot really make sense of this statement ... what exactly is Rotman trying to tell us?Question 3

In the above text Rotman writes the following:

" ... ... Hence \(\displaystyle [x] \in U ( \mathbb{I}_m )\), and so each \(\displaystyle [r] \in U ( \mathbb{I}_m )\) has an inverse in \(\displaystyle U ( \mathbb{I}_m )\) ... ... "

Can someone please explain how this follows ...
Hope someone can help ...Peter

 

[caffeinemachine]

I am reading Joseph J. Rotman's book, Advanced Modern Algebra and am currently focused on Chapter 1: Groups I.

I need some help with the proof of Proposition 1.52.

Proposition 1.52 reads as follows:

I have several related questions that need clarification ...Question 1

In the above text Rotman writes the following:

" ... ... If \(\displaystyle (a,m) = 1\), then \(\displaystyle [a][x] = 1\) can be solved for \(\displaystyle [x]\) in \(\displaystyle \mathbb{I}_m\). ... ... "


Note: Rotman uses \(\displaystyle \mathbb{I}_m\) for the group of integers mod \(\displaystyle m \) and uses \(\displaystyle (a,m)\) for the gcd of \(\displaystyle a\) and \(\displaystyle m\) ...

Can someone explain to me exactly why the above statement by Rotman follows ...Question 2

In the above text Rotman writes the following:

" ... ... Now \(\displaystyle (x,m) = 1\), for \(\displaystyle rx + sm = 1\) for some integer \(\displaystyle s\) and so \(\displaystyle (x,m) = 1\) ... ... "

I cannot really make sense of this statement ... what exactly is Rotman trying to tell us?Question 3

In the above text Rotman writes the following:

" ... ... Hence \(\displaystyle [x] \in U ( \mathbb{I}_m )\), and so each \(\displaystyle [r] \in U ( \mathbb{I}_m )\) has an inverse in \(\displaystyle U ( \mathbb{I}_m )\) ... ... "

Can someone please explain how this follows ...
Hope someone can help ...Peter

To answer your first question: If $\gcd(a, m)=1$, then there exists integers $p$ and $q$ such that $ap+mq=1$. This is quite a standard theorem and is not so hard to prove using inductive reasoning.

Now solving $[a][x]=1$ in $I_m$ is equivalent to finding integers $x$ and $y$ such that $ax+my=1$.

 

[Math Amateur]

To answer your first question: If $\gcd(a, m)=1$, then there exists integers $p$ and $q$ such that $ap+mq=1$. This is quite a standard theorem and is not so hard to prove using inductive reasoning.

Now solving $[a][x]=1$ in $I_m$ is equivalent to finding integers $x$ and $y$ such that $ax+my=1$.

Thanks for the help caffeinemachine ...

But ... ... can you clarify exactly why solving $[a][x]=1$ in $I_m$ is equivalent to finding integers $x$ and $y$ such that $ax+my=1$By the way ... hope someone can help with Questions 2 and 3 in my post above ...

Peter

 

[caffeinemachine]

Thanks for the help caffeinemachine ...

But ... ... can you clarify exactly why solving $[a][x]=1$ in $I_m$ is equivalent to finding integers $x$ and $y$ such that $ax+my=1$By the way ... hope someone can help with Questions 2 and 3 in my post above ...

Peter

Note that for any integer $y$, the symbol $[y]$ as an element of $I_m$ denotes the set of all the integers which leave the same remainder when divided by $m$ as $y$ does. An equivalent way of describing $[y]$ is the set of all integers $z$ such that $m$ divides $y-z$.

Now $I_m$ is nothing but $\{[y]: y\in \mathbf Z\}$. We define a product operation on $I_m$ by declaring $[a]=[ab]$. One needs to check that this is well--defined but this is easy.

Note that $[1]$ is the identity of this product, that is, $[a][1]=[a]$ for all $a$.

Most people prefer writing $1$ instead of $[1]$.

Now $[a][x]=1$ in $I_m$ means $[ax]=[1]$ in $I_m$, That means $m$ divides $ax-1$.

Can you finish?

 

[Math Amateur]

Note that for any integer $y$, the symbol $[y]$ as an element of $I_m$ denotes the set of all the integers which leave the same remainder when divided by $m$ as $y$ does. An equivalent way of describing $[y]$ is the set of all integers $z$ such that $m$ divides $y-z$.

Now $I_m$ is nothing but $\{[y]: y\in \mathbf Z\}$. We define a product operation on $I_m$ by declaring $[a]=[ab]$. One needs to check that this is well--defined but this is easy.

Note that $[1]$ is the identity of this product, that is, $[a][1]=[a]$ for all $a$.

Most people prefer writing $1$ instead of $[1]$.

Now $[a][x]=1$ in $I_m$ means $[ax]=[1]$ in $I_m$, That means $m$ divides $ax-1$.

Can you finish?

Thanks caffeinemachine ...

I think the rest of the proof goes as follows:

\(\displaystyle m \mid ax - 1 \Longrightarrow \exists \ y' \text{ such that } ax - 1 = m y'\)

\(\displaystyle \Longrightarrow ax - my' = 1\)

Now take \(\displaystyle y = -y'\) and we have \(\displaystyle ax + my = 1 \)

and since \(\displaystyle (a, m) = 1\) we know this has a solution for \(\displaystyle x\) ... ... and, indeed \(\displaystyle y\).Is that correct?

Peter

 

[caffeinemachine]

Everything is fine. Except I do not understand what you mean by the following:

and since \(\displaystyle (a, m) = 1\) we know this has a solution for \(\displaystyle x\) ... ... and, indeed \(\displaystyle y\).

What I would say is that solving $[a][x]=1$ in $I_m$ is equivalent to finding $x_0$ and $y_0$ such that $ax_0+by_0=1$. One such an ordered pair $(x_0, y_0)$ is found, the required solution is $[x_0]$.

 

[Math Amateur]

Everything is fine. Except I do not understand what you mean by the following:

What I would say is that solving $[a][x]=1$ in $I_m$ is equivalent to finding $x_0$ and $y_0$ such that $ax_0+by_0=1$. One such an ordered pair $(x_0, y_0)$ is found, the required solution is $[x_0]$.

Thanks again ... most grateful for your help ...

Now I. Am hoping that someone will help with my questions 2 and 3 in my opening post ...

Peter

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Everything is fine. Except I do not understand what you mean by the following:

What I would say is that solving $[a][x]=1$ in $I_m$ is equivalent to finding $x_0$ and $y_0$ such that $ax_0+by_0=1$. One such an ordered pair $(x_0, y_0)$ is found, the required solution is $[x_0]$.

Thanks again ... most grateful for your help ...

Now I. Am hoping that someone will help with my questions 2 and 3 in my opening post ...

Peter

 

[Euge]

There are some typos in the proof of the proposition. Reading through the argument, I suspect $a = r$. Also, in the statement "Now $(x,m) = 1$, for $rx + sm = 1$ for some integer $s$, so $(x,m) = 1$," remove the phrase "so $(x,m) = 1$". This should clear up Question 2. As for Question 3, he proved that every $[r] \in U(\Bbb I_m)$ has an inverse $[x]\in U(\Bbb I_m)$ (i.e., closure under inverses hold in $U(\Bbb I_m)$.