algebraic structures代数的構造だいすうてきこうぞう and binary operations二項演算にこうえんざん: basic exercises

1Corresponding lectures

Problems 1–3 can be attempted after the lecture on binary operations. Attempt Problem 4 after the lecture on semigroups, monoids, and groups.

2Related exercises

3Problem 1: decide whether an operation is binary

Is subtraction[引き算] on $$\mathbb{N}$$\mathbb N a binary operation二項演算にこうえんざん? Give a reason.

3.1Answer

It is not a binary operation二項演算にこうえんざん. For example,

and $$-3\notin \mathbb{N}$$-3\notin\mathbb N.

3.2Explanation

For a binary operation二項演算にこうえんざん $$\text{[PARSE ERROR: Undefined("Command("ast")")]}:S\times S\to S$$\ast:S\times S\to S, the result obtained from any two elements must return to the same set $$S$$S. Here closure閉包性へいほうせい fails.

4Problem 2: find an identity element単位元たんいげん

On $$\mathbb{Z}$$\mathbb Z, consider the operation $$a\text{[PARSE ERROR: Undefined("Command("ast")")]}b=a+b-3$$a\ast b=a+b-3. Find the identity element単位元たんいげん.

4.1Answer

Let the identity element単位元たんいげん be $$e$$e. For every $$a$$a, we need

Therefore

so $$e=3$$e=3. Similarly, $$e\text{[PARSE ERROR: Undefined("Command("ast")")]}a=a$$e\ast a=a also holds, so the identity element単位元たんいげん is 3.

4.2Explanation

An identity element単位元たんいげん is an element that does nothing with respect to the operation. It is not necessarily 0 from ordinary addition. When the operation changes, the identity element単位元たんいげん can also change.

5Problem 3: find an inverse

For the operation in Problem 2, find the inverse element逆元ぎゃくげん of an element $$a$$a.

5.1Answer

Let the inverse be $$b$$b. Since the identity element単位元たんいげん is 3, we need

That is,

so

Since this operation satisfies $$a\text{[PARSE ERROR: Undefined("Command("ast")")]}b=b\text{[PARSE ERROR: Undefined("Command("ast")")]}a$$a\ast b=b\ast a, the opposite order also gives $$b\text{[PARSE ERROR: Undefined("Command("ast")")]}a=3$$b\ast a=3.

5.2Explanation

An inverse is defined relative to the identity element and must work on both sides. Here the identity is 3, and commutativity lets one calculation verify both orders.

6Problem 4: semigroup, monoid, or group?

For the operation $$a\text{[PARSE ERROR: Undefined("Command("ast")")]}b=a+b-3$$a\ast b=a+b-3 from Problems 2 and 3, check closure and associativity, and determine which of the conditions for a semigroup, a monoid, and a group are satisfied by $$(\mathbb{Z},\text{[PARSE ERROR: Undefined("Command("ast")")]})$$(\mathbb Z,\ast).

6.1Answer

For all $$a,b\in \mathbb{Z}$$a,b\in\mathbb Z, we have $$a+b-3\in \mathbb{Z}$$a+b-3\in\mathbb Z, so closure holds. For all $$a,b,c\in \mathbb{Z}$$a,b,c\in\mathbb Z,

Thus the operation is associative and $$(\mathbb{Z},\text{[PARSE ERROR: Undefined("Command("ast")")]})$$(\mathbb Z,\ast) is a semigroup. By Problem 2 its identity is 3, so it is a monoid. By Problem 3, every $$a$$a has the inverse $$6-a\in \mathbb{Z}$$6-a\in\mathbb Z. Therefore $$(\mathbb{Z},\text{[PARSE ERROR: Undefined("Command("ast")")]})$$(\mathbb Z,\ast) is a group.

6.2Explanation

Computing an identity and inverses alone does not yet prove that an operation forms a group. One must separately verify that the rule is a binary operation on the set and that it is associative.