Group Table: Understanding the Identity Element and Avoiding Repeated Values

[joelkato1605]

Homework Statement

Group table

Relevant Equations

n/a

*	s	t	u	v
s	s?	v?	u	v?
t	t	v
u	u?
v	v?

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?

 

[fresh_42]

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

*	s	t	u	v
s	s?	v?	u	v?
t	t	v
u	u?
v	v?

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.

 

[joelkato1605]

Right, do I need to use something like U*T=S*U*T in some way?

 

[fresh_42]

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?

 

[joelkato1605]

So the table should look like this

	s	t	u	v
s	s	t	u	v
t	t	v	s	u
u	u	s	v	t
v	v	u	t	s

 

[joelkato1605]

Thanks for the help.

 

[fresh_42]

So the table should look like this

	s	t	u	v
s	s	t	u	v
t	t	v	s	u
u	u	s	v	t
v	v	u	t	s

What do you get if you put all ##s## on the main diagonal?