3
$\begingroup$

Let $G= \{ 0,1,2,3,4,5,6,7 \}$ and assume $G$ is a group under an operation $*$ with these properties

(i) $a*b \leq a+b$ $\forall a, b \in G$

(ii) $a*a=0$ $\forall a \in G$

write out the operation table for $G$

Its leq im fixing it

hint1: if $ab=ac$ in a group $G$, then $b=c$

hint2: each element of a finite group G appears exactly once in each row and exactly once in each column of the operation table

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From property 2 there will be bunch of zeros going down the diagonal

so far the 8s are just nonsense like a blank

\begin{array}{|c|c|c|c|c|c|c|c|} \hline * & 0& 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 0 & 0 & 8 & 8 & 8 & 8 & 8 & 8 & 8 \\ \hline 1 & 8 & 0 & 8 & 8 & 8 & 8 & 8 & 8 \\ \hline 2 & 8 & 8 & 0 & 8 & 8 & 8 & 8 & 8 \\ \hline 3 & 8 & 8 & 8 & 0 & 8 & 8 & 8 & 8 \\ \hline 4 & 8 & 8 & 8 & 8 & 0 & 8 & 8 & 8 \\ \hline 5 & 8 & 8 & 8 & 8 & 8 & 0 & 8 & 8 \\ \hline 6 & 8 & 8 & 8 & 8 & 8 & 8 & 0 & 8 \\ \hline 7& 8 & 8 & 8 & 8 & 8 & 8 & 8 & 0 \\ \hline \end{array}

for the first row $$ 0*1 \leq 0+1$$

that is for whole row

$$\begin{aligned} 0*1 & \leq 1 \\ 0*2 & \leq 2 \\ 0*3 & \leq 3 \\ 0*4 & \leq 4 \\ 0*5 & \leq 5 \\ 0*6 & \leq 6 \\ 0*7 & \leq 7 \end{aligned} $$

things will fall as $0*7=7, \dots ,0*1=1$

so far the table obtained by the properties seems commutative \begin{array}{|c|c|c|c|c|c|c|c|} \hline * & 0& 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 1 & 0 & 8 & 8 & 8 & 8 & 8 & 8 \\ \hline 2 & 2 & 8 & 0 & 8 & 8 & 8 & 8 & 8 \\ \hline 3 & 3 & 8 & 8 & 0 & 8 & 8 & 8 & 8 \\ \hline 4 & 4 & 8 & 8 & 8 & 0 & 8 & 8 & 8 \\ \hline 5 & 5 & 8 & 8 & 8 & 8 & 0 & 8 & 8 \\ \hline 6 & 6 & 8 & 8 & 8 & 8 & 8 & 0 & 8 \\ \hline 7& 7 & 8 & 8 & 8 & 8 & 8 & 8 & 0 \\ \hline \end{array}

Does this process needs to be repeated 6 more times??? is there some cool way to do it????

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    You must have $1*2=3$, $1*4=5$, $1*6=7$, $2*4=6$, and then everything else follows. – 2017-01-28

1 Answers 1

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Well, it seems a bit like a sudoku :)

I assume, that by $+$ you mean adding modulo 8.

See, that for for every $k$ and $x\leq k$, $x\neq k-x$ we have $$x*(k-x)=k=x+(k-x)$$ We have then:

\begin{array}{|c|c|c|c|c|c|c|c|} \hline * & 0& 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 1 & 0 & 3 & 4 & 5 & 6 & 7 & . \\ \hline 2 & 2 & 3 & 0 & 5 & 6 & 7 & . & . \\ \hline 3 & 3 & 4 & 5 & 0 & 7 & . & . & . \\ \hline 4 & 4 & 5 & 6 & 7 & 0 & . & . & . \\ \hline 5 & 5 & 6 & 7 & . & . & 0 & . & . \\ \hline 6 & 6 & 7 & . & . & . & . & 0 & . \\ \hline 7 & 7 & . & . & . & . & . & . & 0 \\ \hline \end{array} We can now place missing $2$ in row and column for 1, missing 6 in row/column for 3 and missing 5 in row/column for 6 and 7:

\begin{array}{|c|c|c|c|c|c|c|c|} \hline * & 0& 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 1 & 0 & 3 & 4 & 5 & 6 & 7 & 2 \\ \hline 2 & 2 & 3 & 0 & 5 & 6 & 7 & . & . \\ \hline 3 & 3 & 4 & 5 & 0 & 7 & . & . & 6 \\ \hline 4 & 4 & 5 & 6 & 7 & 0 & . & . & . \\ \hline 5 & 5 & 6 & 7 & . & . & 0 & . & . \\ \hline 6 & 6 & 7 & . & . & . & . & 0 & 5 \\ \hline 7 & 7 & 2 & . & 6 & . & . & 5 & 0 \\ \hline \end{array}

See, that (beginning form right) we can easilly fill the row/column for 5 \begin{array}{|c|c|c|c|c|c|c|c|} \hline * & 0& 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 1 & 0 & 3 & 4 & 5 & 6 & 7 & 2 \\ \hline 2 & 2 & 3 & 0 & 5 & 6 & 7 & . & . \\ \hline 3 & 3 & 4 & 5 & 0 & 7 & 1 & . & 6 \\ \hline 4 & 4 & 5 & 6 & 7 & 0 & 2 & . & . \\ \hline 5 & 5 & 6 & 7 & 1 & 2 & 0 & 3 & 4 \\ \hline 6 & 6 & 7 & . & . & . & 3 & 0 & 5 \\ \hline 7 & 7 & 2 & . & 6 & . & 4 & 5 & 0 \\ \hline \end{array}

And missing 2 in row/column for 3: \begin{array}{|c|c|c|c|c|c|c|c|} \hline * & 0& 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 1 & 0 & 3 & 4 & 5 & 6 & 7 & 2 \\ \hline 2 & 2 & 3 & 0 & 5 & 6 & 7 & . & . \\ \hline 3 & 3 & 4 & 5 & 0 & 7 & 1 & 2 & 6 \\ \hline 4 & 4 & 5 & 6 & 7 & 0 & 2 & . & . \\ \hline 5 & 5 & 6 & 7 & 1 & 2 & 0 & 3 & 4 \\ \hline 6 & 6 & 7 & . & 2 & . & 3 & 0 & 5 \\ \hline 7 & 7 & 2 & . & 6 & . & 4 & 5 & 0 \\ \hline \end{array}

In row/column for 2 there are 1 and 4 mising. Because $7*5=4$, it must be $7*2=1$

\begin{array}{|c|c|c|c|c|c|c|c|} \hline * & 0& 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 1 & 0 & 3 & 4 & 5 & 6 & 7 & 2 \\ \hline 2 & 2 & 3 & 0 & 5 & 6 & 7 & 4 & 1 \\ \hline 3 & 3 & 4 & 5 & 0 & 7 & 1 & 2 & 6 \\ \hline 4 & 4 & 5 & 6 & 7 & 0 & 2 & . & . \\ \hline 5 & 5 & 6 & 7 & 1 & 2 & 0 & 3 & 4 \\ \hline 6 & 6 & 7 & 4 & 2 & . & 3 & 0 & 5 \\ \hline 7 & 7 & 2 & 1 & 6 & . & 4 & 5 & 0 \\ \hline \end{array}

Now it is just placing missing 1 in row/column for 6 and missing 3 in row/column for 7: \begin{array}{|c|c|c|c|c|c|c|c|} \hline * & 0& 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 1 & 0 & 3 & 4 & 5 & 6 & 7 & 2 \\ \hline 2 & 2 & 3 & 0 & 5 & 6 & 7 & 4 & 1 \\ \hline 3 & 3 & 4 & 5 & 0 & 7 & 1 & 2 & 6 \\ \hline 4 & 4 & 5 & 6 & 7 & 0 & 2 & \color{red}1 & 3 \\ \hline 5 & 5 & 6 & 7 & 1 & 2 & 0 & 3 & 4 \\ \hline 6 & 6 & 7 & 4 & 2 & \color{red}1 & 3 & 0 & 5 \\ \hline 7 & 7 & 2 & 1 & 6 & 3 & 4 & 5 & 0 \\ \hline \end{array}

But... wait! Why the 1s we just put in this array are glowing red? See, that $$6+4=2>1$$ So there is no operation satisfying your conditions.

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    It is not clear to me why you assumed that addition was modulo $8$, which was not stated in the problem. – 2017-01-28
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    @DerekHolt Just for fun - the problem seems a lot more interesting (and unfortunately leads to that sad conclusion, that there is no solution). If we skip this assumption, after step 2 we can just start filling the table just like a sudoku. – 2017-01-28