Artin Problems: Section 2.2

Problems from Chapter 2: Matrices, Section 2: Groups and Subgroups.

Problem 2.2.1 :

Produce a multiplication table for S_3.

 I’ll leave this one for now as it is very tedious to implement in latex on wordpress with my current knowledge of the platform.

Problem 2.2.2 : 

Let S be a set with an associative law of composition and with identity. Prove that the subset G \subset S consisting of the invertible elements is a group.

S contains the identity, which is it’s own inverse so G contains identity. Each element has an inverse by hypothesis. If a,b \in G then their product ab \in G as the product of inverse is inverse of ab.

Problem 2.2.3 : 

Let x,y,z,w be elements of a group G.

(a) Solve for y, given xyz^{-1}w = 1.

y = x^{-1}zw^{-1}

(b) Suppose that xyz = 1. Does it follow that (i) yzx = 1? Does it follow that (ii) $yxz = 1$?

It does not follow as group laws do not necessarily commute. Note that if yxy^{-1}x^{-1} = 1 then (ii) follows, and if yxy^{-1}x^{-1} = 1 and zxz^{-1}x^{-1} = 1 then (i) follows.

Problem 2.2.4 : 

Determine which of the following pairs are subgroups and which are not.

(a) GL_n(\mathbb{R}) is a subgroup of GL_n(\mathbb{C}). 

(b) \{1,-1\} is a subgroup of \mathbb{R}^\times

(c) The set of positive integers (with addition) is not a subgroup of \mathbb{Z}^+

(d) The set of positive reals (with multiplication) is a subgroup of \mathbb{R}^\times

(e) The set of 2 \times 2 matrices such that a_{11} = a and a_{ij}= 0 otherwise, is not a subgroup of GL_n(\mathbb{R})

Problem 2.2.5 : 

Show that if a subgroup $latex H \subset G$ has an identity, then it must be the identity of G and show that the same statement is also true of inverse.

Suppose H has identity e_H and G has identity e_G. Then for h \in H \subset G we have that $e_Hh = e_Gh = h. With the application of inverse we can show that these must be equal.

Suppose h \in H has inverse h^{-1}_H and G has inverse h^{-1}_G. As identity in H and G are the same element then inverses must be equal as they describe the same map.

Problem 2.2.6 : 

Let G be a group. Define the opposite group (G^\ast) on the same underlying set with law of composition \ast so that $a b = ba. Prove G^\ast is a group.

\ast is a permutation of terms of a group operation and so must be associative as group operations are associative.

G^\ast contains the identity element and a \ast 1 = 1a = a.

 G^\ast contains each inverse element and each left inverse is a right inverse, so inverses exist and are unique.

If a,b \in G then ab = b \ast a and ba = a \ast ba \in G so G^\ast closed under group operation. So G^\ast is a group.

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