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September 22nd, 2018, 06:45 PM  #1 
Newbie Joined: Jun 2018 From: Brasil Posts: 20 Thanks: 0  algebraic structures
how to complete the table? thanks 
September 22nd, 2018, 07:36 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,111 Thanks: 2326 
Hint: + 0 1 2 3 $\ \ $  00 1 2 3 11 2 3 0 22 3 0 1 33 0 1 2 
September 22nd, 2018, 07:57 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,638 Thanks: 1475 
$c \odot a = e$ $a \odot c \odot a = a \odot e = a$ $(a \odot c) \odot a = a$ $a \odot c = e$ $c \odot b = a$ $a \odot c \odot b = a \odot a$ $e \odot b = a \odot a$ $a \odot a = b$ $b \odot a = c$ $a \odot b \odot a = a \odot c$ $(a \odot b) \odot a = e$ $a \odot b = a^{1}$ $a \odot b = c$ $b \odot a = c$ $b \odot a \odot a = c \odot a$ $b \odot b = e$ $b \odot c = a \odot a \odot c$ $b \odot c = a \odot e$ $b \odot c = a$ $c \odot c = c \odot a \odot b$ $c \odot c = e \odot b$ $c \odot c = b$ $\begin{array}{clcccc} & &e &a &b &c \\ \hline \\ e & &e &a &b &c \\ a & &a &b &c &e \\ b & &b &c &e &a \\ c & &c &e &a &b \end{array}$ $a\odot b = b\odot a = c$ $b \odot c = c \odot b = a$ $a \odot c = c \odot a = e$ $G$ is Abelian 
September 23rd, 2018, 02:59 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,111 Thanks: 2326 
Or, as each row and column must contain a permutation of the four elements, 1st step: e a b c a b $\ \ $ e b c c e a b 2nd step: e a b c a b c e b c e a c e a b 

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