UA-AUG-AUMAT229 Discovery guide

Section 5.2 Discovery guide

Discovery 5.1. The full subgroup.

If $G$ is a group, then $G$ itself is a collection of group elements from $G .$ Does $G$ itself satisfy the definition of Subgroup of a group?

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Discovery 5.2. The trivial subgroup.

Suppose $G$ is a group with identity element $e .$ Does the collection ${e}$ satisfy the definition of Subgroup of a group? That is, does the collection ${e}$ satisfy the four group axioms?

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When we studied subspaces of vector spaces in linear algebra, we found that it wasn't necessary to re-check all ten vector space axioms, and we developed the Subspace Test to identify the three conditions that needed to be met in order to be sure that a collection of vectors in a vector space would, on its own, satisfy all ten axioms.

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Similarly, we do not need to check all four group axioms to verify that a collection of group elements forms a subgroup. We may actually reduce verification of a subgroup to just two conditions (though we will begin by reducing to three).

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Discovery 5.3. The Subgroup Test (Version 1).

Recall that the four group axioms are closure, associativity, identity, and inverses. By definition, to be a subgroup, a (non-empty) collection $H$ of group elements from $G$ needs to itself satisfy those four axioms under the same binary operation that makes $G$ a group.

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(a)

Since $G$ is assumed to be a group, the associativity axiom is true for all elements in $g .$ Does associativity need to be checked for elements in $H ?$

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(b)

Suppose $H$ satisfies the closure axiom and the inverses axiom. Convince yourself that the identity of $G$ will be an element of $H,$ so that $H$ also satisfies the identity axiom.

Hint.

Start with an arbitrary element $h$ in $H .$ (We might as well assume that $H$ is non-empty.) First use the assumption that $H$ satisfies the inverses axiom, and then use the assumption that $H$ satisfies the closure axiom.

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In Discovery 5.3 you have justified the following theorem.

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Theorem 5.2.1. The Subgroup Test (Version 1).

Suppose $H$ is a collection of elements from a group $G$ that is

non-empty;
closed under the group operation, and
closed under taking inverses.

Then $H$ is a subgroup of $G .$

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Discovery 5.4.

Suppose $g$ is an element of a group $G .$ Use Theorem 5.2.1 to verify that the Cyclic subgroup generated by $g$ deserves the “subgroup” part of its name.

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Discovery 5.5.

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(a)

Recall that $Q^{pos}$ is the multiplicative group of positive rational numbers. What sort of the elements does the cyclic subgroup $⟨ 2 ⟩$ contain? Give some examples.

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(b)

In the additive group $Z,$ what sort of elements does the cyclic subgroup $⟨ 2 ⟩$ contain? Give some examples.

This subgroup is usually denoted $2 Z$ — do you see why?

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Discovery 5.6. The trivial generator.

Suppose $G$ is a group with identity element $e .$ What elements does the cyclic subgroup $⟨ e ⟩$ contain?

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Discovery 5.7.

Sometimes an entire group is cyclic — determine two different integers $a$ so that $Z = ⟨ a ⟩$ is true.

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Discovery 5.8. Abelian versus non-Abelian.

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(a)

Is it possible for an Abelian group to have a non-Abelian subgroup?

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(b)

Is it possible for a non-Abelian group to have an Abelian subgroup?

Hint.

Consider cyclic subgroups.

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Discovery 5.9.

Use Theorem 5.2.1 to verify that the intersection of two subgroups is always a subgroup. (The intersection of two collections is made up of those elements that are common to both.)

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Discovery 5.10. The Subgroup Test (Version 2).

Again, assume that $H$ is a (non-empty) collection of elements from a group $G,$ but this time assume that $H$ satisfies a different kind of closure axiom: whenever $x, y$ are elements of $H,$ so is $x y^{- 1}$ .

Verify that this condition implies that $H$ will pass the conditions of Theorem 5.2.1 (and so will be a subgroup of $G$ ).

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In Discovery 5.10 you have justified the following theorem.

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Theorem 5.2.2. The Subgroup Test (Version 2).

Suppose $H$ is a non-empty collection of elements from a group $G$ so that whenever $x$ and $y$ are elements of $H,$ so is the result of computing $x y^{- 1} .$ Then $H$ is a subgroup of $G .$

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Our analysis of dihedral groups in Topic 4 shows that

D_{n} = ⟨ r, s ⟩,

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where

r

is rotation by

2 π / n

about the centre of a face of the regular

n

-agon and

s

is rotation by

π

about some chosen axis either through two opposing vertices (in case

n

is even) or through a vertex and the centre of the opposing edge (in case

n

is even). But all of these dihedral groups are finite.

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The last activity concerns what could be considered an infinite dihedral group: a group consisting of certain kinds of transformations of an infinitely long line with regularly-spaced vertices.

A “regular $\infty$-agon.” — 🔗
Figure 5.2.3. A “regular $\infty$ -agon.”

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Discovery 5.11.

An isometry is a transformation that preserves distances between two points. That is, every pair of points will be the same distance apart after an isometry transformation as they were before the transformation.

Let $D_{\infty}$ be the group of all isometries of the line in Figure 5.2.3 that map vertices to vertices, where the group operation is, as usual, composition of transformations.

Think about what kind of transformations $D_{\infty}$ contains. Then come up with two specific transformations, $r$ and $s$ so that

D_{\infty} = ⟨ r, s ⟩ .

Hint.

If you want to work with the analogy between this $D_{\infty}$ and the finite dihedral groups $D_{n},$ it is possible to consider both $r$ and $s$ as “rotations” that are similar in nature to the usual $r$ and $s$ rotations in $D_{n}$ if you picture wrapping the “ends” of the line in Figure 5.2.3 around together to meet at a “point at infinity” (though there is no actual vertex there), creating a shape that you could take to be a regular polygon with an infinite number of sides.