cayley's table

[Frankenstein143]

Can someone explain to me, why is a*a= e?
It is a Cayley's table. e is defined as an identity element/neutral element. By multiplication is e:=1
If e=1, then why is a*a=e?
I am confused because if I want to get e, I have to do this: a*1/a=e

In the next picture is a*a=b why?

 

[Dr.Peterson]

Can someone explain to me, why is a*a= e?
It is a Cayley's table. e is defined as an identity element/neutral element. By multiplication is e:=1
If e=1, then why is a*a=e?
I am confused because if I want to get e, I have to do this: a*1/a=e

In the next picture is a*a=b why?

View attachment 27362

You may need to show context, but it appears that this table DEFINES the operation *; it is not just ordinary multiplication. So each table just tells you the facts listed below the table. You can fill in a table any way you want to, and it will define an operation (which may or may not be an interesting or useful operation!).

I want to see what was said before each table was shown, and what you were taught about Cayley tables and operations.

EDIT: In particular, I want to know whether this table was just given as an example of an operation, or you are filling it in according to some requirement, such as having been told it has to represent a group. What have you learned so far?

 

[lex]

[MATH]a \cdot a[/MATH] can't be anything but [MATH]e[/MATH]. The only other option is that it is [MATH]a[/MATH]. If [MATH]a \cdot a[/MATH] is [MATH]a[/MATH], then [MATH]a[/MATH] is the identity element.
Also there is only one group of order 2.

Again (for the same reason as above) [MATH]a \cdot a[/MATH] can only be [MATH]b[/MATH] or [MATH]e[/MATH]. If it is [MATH]e[/MATH], then [MATH]a[/MATH] is an element of order 2 which is impossible in a group of order 3. (The order of the element must divide the order of the group).
Also, there is only one group of order 3.

 

[Steven G]

1/3 is not 1, 3 is not 1; yet 1/3 * 3 is 1!

Here is a better example. -1 is not 1, yet -1*-1 is 1!
Make up a Cayley table, using regular multiplication with only 1 and -1 as elements. Is this set with regular multiplication a group??