groups群ぐん and subgroups部分群ぶぶんぐん: basic exercises

1Corresponding lectures

2Suggested order

After “Semigroups, monoids, and groups,” complete Problem 1 and the proof exercise. After “Subgroups and generators,” complete Problems 2–3. The page prerequisites describe what is needed to complete the whole page.

3Related exercises

4Problem 1: test the semigroup, monoid, and group conditions

Here $$\mathbb{N}=\{0,1,2,\dots \}$$\mathbb N=\{0,1,2,\ldots\}. Determine which of the semigroup, monoid, and group群ぐん conditions are satisfied by $$(\mathbb{N},+)$$(\mathbb N,+).

4.1Answer

Addition is closed and associative on the natural numbers, so $$(\mathbb{N},+)$$(\mathbb N,+) is a semigroup. Since 0 is an identity element, it is also a monoid. It is not a group, because, for example, the additive inverse $$-3$$-3 of 3 is not in $$\mathbb{N}$$\mathbb N.

4.2Explanation

In a group, every element must have an inverse. Closure and associativity結合法則けつごうほうそく alone are not enough.

5Problem 2: decide whether it is a subgroup部分群ぶぶんぐん

Is $$2\mathbb{Z}$$2\mathbb Z a subgroup部分群ぶぶんぐん of $$(\mathbb{Z},+)$$(\mathbb Z,+)?

5.1Answer

It is a subgroup部分群ぶぶんぐん. It is nonempty, and for any $$2a,2b\in 2\mathbb{Z}$$2a,2b\in2\mathbb Z,

5.2Explanation

For an additive group, the subgroup部分群ぶぶんぐん test can be checked by closure閉包性へいほうせい under differences.

6Problem 3: generated subgroup部分群ぶぶんぐん

In $$(\mathbb{Z}/8\mathbb{Z},+)$$(\mathbb Z/8\mathbb Z,+), find the subgroup部分群ぶぶんぐん generated by $$\left[2\right]$$[2].

6.1Answer

Every integer multiple of $$\left[2\right]$$[2] has the form $$k\left[2\right]=\left[2k\right]$$k[2]=[2k]. Write any $$k\in \mathbb{Z}$$k\in\mathbb Z as $$k=4q+r$$k=4q+r with $$r\in \{0,1,2,3\}$$r\in\{0,1,2,3\}. Then

so every integer multiple is one of $$\left[0\right],\left[2\right],\left[4\right],\left[6\right]$$[0],[2],[4],[6]. Conversely, taking $$q=0$$q=0 and $$k=r=0,1,2,3$$k=r=0,1,2,3 gives these four elements, respectively. Both inclusions therefore hold, and

6.2Explanation

The subgroup部分群ぶぶんぐん generated by a specified element is the set of all elements reachable from it using the operation and inverses.

7Proof exercise: inverse of a product and equations

7.1Problem

For elements $$a,b$$a,b of a group $$G$$G, prove $$(ab{)}^{-1}=b^{-1}{a}^{-1}$$(ab)^{-1}=b^{-1}a^{-1}. Also solve $$a\text{[PARSE ERROR: Undefined("Command("ast")")]}x=b$$a\ast x=b and $$x\text{[PARSE ERROR: Undefined("Command("ast")")]}a=b$$x\ast a=b, and prove that each solution is unique.

7.2Answer

We have

Uniqueness of inverses therefore gives $$(ab{)}^{-1}=b^{-1}{a}^{-1}$$(ab)^{-1}=b^{-1}a^{-1}.

Operating on the left of $$a\text{[PARSE ERROR: Undefined("Command("ast")")]}x=b$$a\ast x=b by $$a^{-1}$$a^{-1} gives $$x=a^{-1}\text{[PARSE ERROR: Undefined("Command("ast")")]}b$$x=a^{-1}\ast b, so every solution is forced to have this value. Operating on the right of $$x\text{[PARSE ERROR: Undefined("Command("ast")")]}a=b$$x\ast a=b by $$a^{-1}$$a^{-1} gives $$x=b\text{[PARSE ERROR: Undefined("Command("ast")")]}{a}^{-1}$$x=b\ast a^{-1}, again forcing every solution to have that value. Substitution verifies both candidates, so each solution exists and is unique.

7.3Explanation

Order matters in a noncommutative group. Undoing a product requires inverses in reverse order, and solving an equation requires operating on the side determined by the position of the unknown.