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 May 17th, 2017, 01:27 PM #1 Newbie   Joined: May 2017 From: Neverland Posts: 8 Thanks: 0 [Set Theory] - filling tablet (G, *) is a group of 4 elements, a,b,c,e are different elements of G and e is a neutral element in (G, *). given this equation: a*b = c*c how can i start filling the tablet of (G, *) ? i've understood that the column and row of the neutral element are just the same as the one that goes with it, a*e=a same goes to b,c and e. and then i have this " every element in a tablet of a group can show only once in each row or column. and can i assume that if a*b = c*c then b*a = c*c as well? i'll appreciate any help or clues
May 17th, 2017, 02:22 PM   #2
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Quote:
 Originally Posted by Adm (G, *) is a group of 4 elements, a,b,c,e are different elements of G and e is a neutral element in (G, *). given this equation: a*b = c*c how can i start filling the tablet of (G, *) ? i've understood that the column and row of the neutral element are just the same as the one that goes with it, a*e=a same goes to b,c and e. and then i have this " every element in a tablet of a group can show only once in each row or column.
Get started by writing down the elements that you know.

Quote:
 Originally Posted by Adm and can i assume that if a*b = c*c then b*a = c*c as well?
No, the group isn't Abelian unless the problem says it is. Now in some cases we can assume commutativity. For example if $xx^{-1} = e$ then we can conclude that $x^{-1}x = e$ as well. But in general you can't assume commutativity.

 May 17th, 2017, 03:27 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 There are only two groups, up to isomorphism, of order 4, the rotation group and Klein four group and they are both Abelian.
May 18th, 2017, 04:43 AM   #4
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Quote:
 Originally Posted by Maschke Get started by writing down the elements that you know.
maybe.jpg

[attached a picture]

so i've put all the Given one's first, then i start like this:
c*a can't be equal to a or c, because that'll make a or c nuetral elements which is wrong (Given that e is nuetral), niether can be a*b cuz it's already exist in the row so c*a = b.

and then c*b = a, because it's the only element left (no element duplicating)

and a*c = b because, a already exists in the row so it can't be, niether a*b or c (no duplicating) so we're left with b.

then can i assume that c*c = a*b = e (nuetral element) ?
because Given that G only Has 4 elements and e is the nuetral one within them, and it's the only one left in this row.

is that's the way ?

 May 21st, 2017, 06:16 AM #5 Newbie   Joined: May 2017 From: Neverland Posts: 8 Thanks: 0 may anyone look up my answer please
May 21st, 2017, 08:49 AM   #6
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Could you notate what you've got? Does't have to be fancy.

 May 26th, 2017, 12:48 PM #7 Newbie   Joined: May 2017 From: Neverland Posts: 8 Thanks: 0 thanks guys i've got it in the end !

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