# 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:

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

Now you prove your claim about the right inverse.

Your proof of (c) is almost correct, yet so far you've only dealt with right inverses, so you must write (and show!) this...and then what axiom of group theory isn't fulfilled?

• In (a) my proof is wrong but still it is associative? I thought |(|a|b)|c = |ab|c and |a||b|c = |ab|c. Why? – MODULUS Oct 27 '14 at 14:55
• Yes, it is associative, yet you didn't prove it. Try to read carefully and understanding my answer. – Timbuc Oct 27 '14 at 14:58
• The way I prove is just like your answer. It is so very long and tedious to type so I just put the final answer. But I was confused that it stops on ||a|b|c = |a||b|c can it still be |ab|c= |ab|c? Also in (c) I use the G1,G2,and G3 axioms but all I want is that, is it correct that I may conclude that it is not a group because both 1/2 and and -1/2 are right inverse of 2. – MODULUS Oct 27 '14 at 15:06
• $$|\,|a|\,b\,| =|\,|a|\,|\cdot|b|=|a|\,|b|$$ , and yes: from the axioms of group theory we know inverses are unique. – Timbuc Oct 27 '14 at 15:52