Group Table: Understanding the Identity Element and Avoiding Repeated Values

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SUMMARY

The discussion focuses on understanding the identity element in group tables and the necessity of avoiding repeated values in rows and columns. It establishes that if s*u=u, then s is the identity element, which leads to the conclusion that all four elements must appear in each row and column of the group table. The participants clarify that blank spaces cannot be left unfilled, as every multiplication must be defined, and they provide a structured approach to filling in the group table correctly.

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Homework Statement
Group table
Relevant Equations
n/a
*stuv
ss?v?uv?
ttv
uu?
vv?

Since s*u=u does that mean s is the identity element? Then I know there can't be repeated values in a row or column so I need to us that to somehow fill in the rest of the blank spaces?
 

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joelkato1605 said:
Homework Statement:: Group table
Relevant Equations:: n/a

*stuv
ss?v?uv?
ttv
uu?
vv?

Since s*u=u does that mean s is the identity element?
Yes, but then s*t=s*v=v cannot be!
Then I know there can't be repeated values in a row or column ...
Right.
... so I need to us that to somehow fill in the rest of the blank spaces?
No. You have to fill them with an element, since every multiplication has to be defined, and blank is no group element.

Hint: There are two possible solutions.
 
Right, do I need to use something like U*T=S*U*T in some way?
 
You have to decide, which element is the identity. Seems, that ##s## has this role. So ##s\cdot a= a## for any other group element. This gives you the first row and first column.

Before you check associativity, look whether you can fill up the remaining ##9## places according to the rule: all ##4## elements must occur in each row and each column! Do you know why there is such a rule?
 
So the table should look like this
stuv
sstuv
ttvsu
uusvt
vvuts
 
Thanks for the help.
 
joelkato1605 said:
So the table should look like this
stuv
sstuv
ttvsu
uusvt
vvuts
What do you get if you put all ##s## on the main diagonal?
 

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