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Say that $S$ is a monoid. Define a subset of $S$ by $S^\times := \{a\in S\mid a \text{ has an inverse}\}$

How can we show that $S^\times$ is a group with the same operation? Can we use this to prove that $U_n$ is a group?

$U_n:= \{a\in \mathbb{Z}_n\mid \gcd(a,n)=1\}$

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    You dont have to do anything to prove that $S^\times$ is a group. A group is by definition a monoid such that every element has an inverse. Since you are "tossing out" everything that does not have an inverse, you are done by contruction. I dont know what you mean by $U_n$.2012-07-06
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    Show that $S^\times$ is closed under the operation and contains the identity element ... as the operation is associative and elements of $S^\times$ have inverses, you have a group. What is $U_n$?2012-07-06
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    **Hint**: can you write $(ab)^{-1}$ in terms of $a^{-1}$ and $b^{-1}$?2012-07-06
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    You should show where you are stuck and what you have tried or at the very least demonstrate that you understand the terms involved in the question (that you know what a group is, for example)2012-07-06
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    Yes, but that is using a more general result to prove one specific instance. What you need to do is show that $U_n$ is closed under multiplication, first. One technical quibble: the "$a$" in $\Bbb{Z}_n$, and in "$\text{gcd}(a,n) = 1$" are not the same kind of thing: one is a congruence class, and another is an integer. You should also *prove* that $[a] \in \Bbb{Z}_n$ is invertible if and only if $\text{gcd}(a,n) = 1$.2012-07-06

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