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September 22nd, 2018, 06:45 PM   #1
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Unhappy algebraic structures

how to complete the table?

thanks
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September 22nd, 2018, 07:36 PM   #2
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Hint:
+ 0 1 2 3
$\ \ $ -------
0|0 1 2 3
1|1 2 3 0
2|2 3 0 1
3|3 0 1 2
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September 22nd, 2018, 07:57 PM   #3
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$c \odot a = e$
$a \odot c \odot a = a \odot e = a$
$(a \odot c) \odot a = a$
$a \odot c = e$

$c \odot b = a$
$a \odot c \odot b = a \odot a$
$e \odot b = a \odot a$
$a \odot a = b$

$b \odot a = c$
$a \odot b \odot a = a \odot c$
$(a \odot b) \odot a = e$
$a \odot b = a^{-1}$
$a \odot b = c$

$b \odot a = c$
$b \odot a \odot a = c \odot a$
$b \odot b = e$

$b \odot c = a \odot a \odot c$
$b \odot c = a \odot e$
$b \odot c = a$

$c \odot c = c \odot a \odot b$
$c \odot c = e \odot b$
$c \odot c = b$

$\begin{array}{clcccc}
&| &e &a &b &c \\
\hline \\
e &| &e &a &b &c \\
a &| &a &b &c &e \\
b &| &b &c &e &a \\
c &| &c &e &a &b
\end{array}$

$a\odot b = b\odot a = c$
$b \odot c = c \odot b = a$
$a \odot c = c \odot a = e$

$G$ is Abelian
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September 23rd, 2018, 02:59 AM   #4
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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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