# Studying for Abstract Algebra: Group

Help me understand more the example of the book so I may understand the whole thing. I am more on detailed-solution-kind of student to make myself get the idea. I am new to this topic and I find Abstract Algebra very difficult to understand and our professor doesn't help me understand but lead me to confusion. That's why I am trying my self to understand all the topics with the help of others who are passionate to this course.

Example 1.2.24 Let ℝ* be the set of all real number except 0. Define * by letting a * b = |a| b

a.)Show that * gives and associative binary operation on ℝ*.

Answer: |ab|c = |ab|c so it is associative.

b.)Show that there is a left identity for * and a right inverse for each element in ℝ*

Answer: Left identity element is 1 and the right inverse is 1/|a|.

c.)Is ℝ* with this binary operation a group?

Answer: It is not a group because both 1/2 and and -1/2 are right inverse of 2.

Your proof of (a) is wrong. You must show that

$$(a*b)*c=a*(b*c)\iff (|a|b)*c = |a|(b*c)\iff \left|\,|a|b\,\right|c=|a||b|c$$

Well, now prove the last equality.

As for (b): why and why? You must prove that. For example:

$$\forall\,a\in\Bbb R^*\;,\;\;1*a=|1|a=a\implies 1\;\;\text{is a left identity}$$

• $$|\,|a|\,b\,| =|\,|a|\,|\cdot|b|=|a|\,|b|$$ , and yes: from the axioms of group theory we know inverses are unique. – Timbuc Oct 27 '14 at 15:52