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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 $$(a,m) = 1$$, then $$[a][x] = 1$$ can be solved for $$[x]$$ in $$\mathbb{I}_m$$. ... ... "
Note: Rotman uses $$\mathbb{I}_m$$ for the group of integers mod $$ m $$ and uses $$(a,m)$$ for the gcd of $$a$$ and $$m$$ ...
Can someone explain to me exactly why the above statement by Rotman follows ...Question 2
In the above text Rotman writes the following:
" ... ... Now $$(x,m) = 1$$, for $$rx + sm = 1$$ for some integer $$s$$ and so $$(x,m) = 1$$ ... ... "
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 $$[x] \in U ( \mathbb{I}_m )$$, and so each $$[r] \in U ( \mathbb{I}_m )$$ has an inverse in $$U ( \mathbb{I}_m )$$ ... ... "
Can someone please explain how this follows ...
Hope someone can help ...Peter
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 $$(a,m) = 1$$, then $$[a][x] = 1$$ can be solved for $$[x]$$ in $$\mathbb{I}_m$$. ... ... "
Note: Rotman uses $$\mathbb{I}_m$$ for the group of integers mod $$ m $$ and uses $$(a,m)$$ for the gcd of $$a$$ and $$m$$ ...
Can someone explain to me exactly why the above statement by Rotman follows ...Question 2
In the above text Rotman writes the following:
" ... ... Now $$(x,m) = 1$$, for $$rx + sm = 1$$ for some integer $$s$$ and so $$(x,m) = 1$$ ... ... "
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 $$[x] \in U ( \mathbb{I}_m )$$, and so each $$[r] \in U ( \mathbb{I}_m )$$ has an inverse in $$U ( \mathbb{I}_m )$$ ... ... "
Can someone please explain how this follows ...
Hope someone can help ...Peter