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Show that the set of complex numbers $z$ with $|z|=1$ is not a group under the operation $*$ denoted by $z_1 * z_2 = |z_1|\times z_2.$

By solving this I found many left identity elements and couldn't find an unique right identitiy. But can't find a particular identity element. Is it enough to say that the set is not a group ?

Also please tell me If a set is satisfying left axioms, but not satisfying right axioms, then we will call the set a group or not ? How many identity element can be in a group ?

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  • $\begingroup$ There is more than one set of axioms for group theory. Your talk of "left axioms" implies that you're using a less common one than average. Please would you let us know, too, how you got your current results. You could be closer than you think! $\endgroup$
    – Shaun
    Sep 25 '19 at 23:41
  • $\begingroup$ @Shaun sorry sir, but I can't find my answers there. I can't understand what that is written in that post, most of them are in codings. $\endgroup$
    – Rajat Dash
    Sep 25 '19 at 23:41
  • $\begingroup$ My link above is to a meta site question on how to use $\LaTeX$ to format mathematics on the main site. $\endgroup$
    – Shaun
    Sep 25 '19 at 23:43
  • $\begingroup$ @Shaun please see this ez=|e|z, here the identity element e can be any member of the set. But if ze=|z|e , then identity element can be only z itself. In first case the identity element is left identity and in second case is right identity. $\endgroup$
    – Rajat Dash
    Sep 25 '19 at 23:45
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    $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$
    – Shaun
    Sep 26 '19 at 0:01
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Since $\lvert z\rvert=1$ for all $z$ in your set, we have $z_1\ast z_2=z_2$ for all $z_1, z_2$ in the set. This violates the Latin square property of group multiplication. Hence it is not a group.


Semigroup theory might be of interest to you.

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