Table of Content
A structural property of a binary structure is one that must be shared by any isomorphic structure.
e.g. (structural)
isomorphism map 過去之後,仍然保留的性質。
e.g. (non-structural)
Problem
Show that the property that the binary operator is commutative is a structure property.
Solution
To prove the property, we have to show that if \((S, *)\) satisfies \(*\) is commutative, then for any \((S', *')\) isomorphic to \((S, *)\), it also satisfies that \(*'\) is commutative.
Assume \(*\) is commutative on \((S, *)\). Given \((S', *')\) isomorphic to \((S, *)\), there exists a isomorphism \(\phi: S \to S'\). Since \(\phi\) is onto, for any \(a', b' \in S'\), there exists \(a, b \in S\) such that \(\phi(a) = a', \phi(b) = b'\), respectively.
Then we have
\[a' *' b' = \phi(a) *' \phi(b) = \phi(a*b) = \phi(b*a) = \phi(b) *' \phi(a) = b' *' a'.\]Hence \(*'\) is commutative on \((S', *')\), which implies the binary operator is commutative is a structure property. ◼
Let \((S, *)\) be a binary structure. An element \(e\) of \(S\) is an identity element for \(*\) if \(e * s = s * e = s\) for all \(s \in S\).
e.g.
A binary structure \((S, *)\) has at most one identity element. That is, if there is an identity element, then it is unique.
Proof(反證法,略)
Suppose that \((S, *)\) has an identity element \(e\) for \(*\). If \(\phi: S \to S'\) is an isomorphism of \((S, *)\) with \((S', *')\), then \(\phi(e)\) is an identity element for \(*'\).
Proof
We need to show that
\[\phi(e) *' s' = s' *' \phi(e) = s'\]for all \(s' \in S'.\)
用 isomporhism 的性質。
A group \((G, *)\) is a set \(G\), with a binary operation \(*\), such that
- \(*\) is associative.
- There exists an identity element \(e \in G\) for \(*\)
- There is an inverse \(a'\) of \(a\) such that \(a' * a = a * a' = e\), for all \(a \in G\).
e.g. (a group)
第三例是 entry 為實數的 \(m \times n\) matrices。
e.g. (not a group)
再舉一個大例子:
The set \(\text{GL}(n, \mathbb{R})\) of all invertible \(n \times n\) matrices under matrix multiplication is a group. (\(\text{GL}\) stands for general linear.)
要驗證 \(\text{GL}\) 是不是 group,除了定義的三個條件之外,還要確認 \(\text{GL}\) 是不是 closed under multiplication!(因為 \(\text{GL}\) 只是某 binary structure 的子集)
利用 \(\text{det}(AB) = \text{det}(A)\text{det}(B)\) 驗證。
A group \((G, *)\) is abelian if its binary operation is commutative.
\(+\) 保留給 commutative binary operation,其他用 \(\cdot\) (multiplication)。(Notation)
Suppose \(G\) is a group.
\(ab = ac \implies b = c\), and \(ba = ca \implies b = c\), for all \(a, b, c \in G\).
用 \(a^{-1}\) 和 \(e\) 證明。
\(ax = b\) and \(ya = b\) have unique solutions \(x\) and \(y\) in \(G\).
Proof \(\rm I\) (existence)
Let \(x = a^{-1}b\) …
Proof \(\rm II\) (uniqueness)
Suppose \(ax_1 = b = ax_2\) …
\(a^{-1}\) and \(e\) is unique, for \(a \in G\).
基本上 uniqueness 都是用反證法。
Corollary
For all \(a, b \in G\), \((ab)^{-1} = b^{-1}a^{-1}\).