description/proof of that for ring, subset of invertible elements forms multiplicative group
Topics
About: ring
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of ring.
Target Context
- The reader will have a description and a proof of the proposition that for any ring, the subset of the multiplicatively-invertible elements forms a multiplicative group.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(S\): \(= \{r \in R \vert \exists r^{- 1} \in R\}\) with the multiplication induced from \(R\)
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Statements:
\(S \in \{\text{ the groups }\}\)
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2: Note
\(R\) itself is an additive Abelian group, by the definition of ring.
But \(R\) itself is not any group under the multiplication in general.
\(S\) is not a group under the addition any more in general.
For example, let \(R = \mathbb{Z}\), then, \(S = \{- 1, 1\}\), and \(S\) is not any group under the addition, because \(1 + 1 \notin S\), for example.
3: Proof
Whole Strategy: Step 1: see that \(S\) satisfies the conditions to be a group.
Step 1:
Let \(s_1, s_2, s_3 \in S\) be any.
\(S\) is closed under the multiplication, because \((s_1 s_2)^{- 1} = {s_2}^{- 1} {s_1}^{- 1}\), because \({s_2}^{- 1} {s_1}^{- 1} s_1 s_2 = {s_2}^{- 1} ({s_1}^{- 1} s_1) s_2 = {s_2}^{- 1} i s_2 = {s_2}^{- 1} s_2 = i\) and \(s_1 s_2 {s_2}^{- 1} {s_1}^{- 1} = s_1 (s_2 {s_2}^{- 1}) {s_1}^{- 1} = s_1 i {s_1}^{- 1} = s_1 {s_1}^{- 1} = i\).
1) \((s_1 \bullet s_2) \bullet s_3 = s_1 \bullet (s_2 \bullet s_3)\): because it holds on the ambient \(R\).
2) \(i \in S\) (called 'identity element') such that \(i \bullet s_1 = s_1 \bullet i = s_1\): \(i \in S\), because \(i^{- 1} = i \in R\), because \(i i = i\), and \(i \bullet s_1 = s_1 \bullet i = s_1\), because it holds on the ambient \(R\).
3) \(s_1^{- 1} \in S\) (called 'inverse element of \(s_1\)') such that \(s_1^{- 1} \bullet s_1 = s_1 \bullet s_1^{- 1} = i\): \(s_1^{- 1} \in S\), because \({s_1^{- 1}}^{- 1} = s_1 \in S\), because \(s_1 s_1^{- 1} = s_1^{- 1} s_1 = i\), and \(s_1^{- 1} s_1 = s_1 s_1^{- 1} = i\), because it holds on the ambient \(R\).
So, \(S\) is a group under the multiplication.
