description/proof of that for group, nonempty subset is subgroup iff it is closed under operation and inversion
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of group.
Target Context
- The reader will have a description and a proof of the proposition that for any group, any nonempty subset is a subgroup if and only if it is closed under the operation and the inversion.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(S\): \(\subseteq G\) such that \(S \neq \emptyset\)
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Statements:
\(S \in \{\text{ the subgroups of } G\}\)
\(\iff\)
\(\forall s_1, s_2 \in S (s_1 s_2 \in S \land {s_1}^{- 1} \in S)\)
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2: Note
The point is that only \(s_1 s_2 \in S\) or only \({s_1}^{- 1} \in S\) is not enough for \(S\) to be a subgroup.
For example, let \(G = \mathbb{Z}\) as the additive group and \(S = \mathbb{N}\), then, \(S\) is closed under the operation, but \(S\) is not any subgroup, because the inverse of \(1\) is not contained in \(S\), for example; let \(G = \mathbb{Z}\) as the additive group and \(S = \{- 1, 1\}\), then, \(S\) is closed under the inversion, but \(S\) is not any subgroup, because \(1 + 1 \notin S\).
3: Proof
Whole Strategy: Step 1: suppose that \(S\) is a subgroup; Step 2: see that \(\forall s_1, s_2 \in S (s_1 s_2 \in S \land {s_1}^{- 1} \in S)\); Step 3: suppose that \(\forall s_1, s_2 \in S (s_1 s_2 \in S \land {s_1}^{- 1} \in S)\); Step 4: see that \(S\) is a subgroup.
Step 1:
Let us suppose that \(S\) is a subgroup.
Step 2:
\(\forall s_1, s_2 \in S (s_1 s_2 \in S \land {s_1}^{- 1} \in S)\) is obvious, because \(S\) is a group.
Step 3:
Let us suppose that \(\forall s_1, s_2 \in S (s_1 s_2 \in S \land {s_1}^{- 1} \in S)\).
Step 4:
Let us see that \(S\) satisfies the conditions to be a group.
As \(\forall s_1, s_2 \in S (s_1 s_2 \in S)\), the operation induced by the operation on \(G\) is well defined on \(S\).
Let \(s_1, s_2, s_3 \in S\) be any.
1) \((s_1 \bullet s_2) \bullet s_3 = s_1 \bullet (s_2 \bullet s_3)\): because it holds on the ambient \(G\).
2) \(i \in S\) (called 'identity element') such that \(i \bullet s_1 = s_1 \bullet i = s_1\): as \(S \neq \emptyset\), there is an \(s \in S\), but \(s^{- 1} \in S\) and \(s s^{- 1} = i \in S\), and \(i s_1 = s_1 i = s_1\), because it holds on the ambient \(G\).
3) \({s_1}^{- 1} \in S\) (called 'inverse element of \(s_1\)') such that \({s_1}^{- 1} \bullet s_1 = s_1 \bullet {s_1}^{- 1} = i\): \({s_1}^{- 1} \in S\), and \({s_1}^{- 1} s_1 = s_1 {s_1}^{- 1} = i\), because it holds on the ambient \(G\).
