Show that $(S,*)$ is a group,where $S=\left\{a,b,c,e\right\}$

Question

Given a set $S=\left\{a,b,c,e\right\}$ equipped with a binary operation $*:S \times S \to S$,such that $$a^2=b^2=c^2=abc=e.$$

Show that based on the table

$$ \begin{array}{r|rr} *&e&a&b&c&\\\hline e&e&a&b&c\\ a&\;a&e&c&b\\ b&\;b&c&e&a\\ c&\;c&b&a&e \end{array} $$

$(S,*)$ is an Abelian group.

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$S$ is closed under $*$ since the elements on each raw are members contained in $S$, every element is it's own inverse and the identity element is $e$, moreover, $*$ is commutative since the table is symmetric along the main diagonal.

I have two questions:

• How to show that $*$ is associative, I checked and there are $\binom{4}{1} \binom{3}{1} \binom{2}{1} =24$ cases to be checked and this is frustrating, so does there exist a better way?
• If I have only the presentation $a^2=b^2=c^2=abc=e$, then how should I draw the Cayley table? I only could form the following table:

$$ \begin{array}{r|rr} *&e&a&b&c&\\\hline e\\ a&\;&e&c&b\\ b&\;&c&e&a\\ c&\;&b&a&e \end{array} $$

How to fill the others with the given presentation?

Answer 1

You can construct a bijection: $$ \varphi:S\rightarrow\mathbb{Z}_2\times\mathbb{Z}_2 $$ defined by \begin{align} \varphi(e)&:=(0,0),\\ \varphi(a)&:=(1,0),\\ \varphi(b)&:=(0,1),\\ \varphi(c)&:=(1,1). \end{align} Now use the Cayley table of $\mathbb{Z}_2\times\mathbb{Z}_2$ to write the Cayley table of $S$.

These tables are the "same", in the sense that if an element $x\in S$ appears in the entry $(i,j)$ of the Cayley table of $S$, then $\varphi(x)\in\mathbb{Z}_2\times\mathbb{Z}_2$ is in the entry $(i,j)$ of the Cayley table of $\mathbb{Z}_2\times\mathbb{Z}_2$. This shows that $S$ is a group.

$S$ is abelian because $\mathbb{Z}_2\times\mathbb{Z}_2$ is such.

Answer 2

We can check associativity of the Cayley table more easily as follows, see

Is there an easy way to see associativity or non-associativity from an operation's table?

"Direct verification of the associativity of a binary operation specified by a Cayley table is cumbersome and tedious. Light's associativity test greatly simplifies the task."

Answer 3

If you only have \begin{array}{r|rr} &e&a&b&c&\\\hline e\\ a&\;&e&c&b\\ b&\;&c&e&a\\ c&\;&b&a&e \end{array} you can complete it uniquely, because in the Cayley table for a group no element can appear twice in a row or in a column.

You can get this table from the data $a^2=b^2=c^2=abc=e$, because if you want to get a group, from $c^2=e$ and $abc=e$ you must get $ab=c$; similarly, from $a^2=abc=e$ you get $bc=a$ (division in groups is unique).

Now you want a group, so it must have a neutral element and only $e$ can be. so you can fill in the table \begin{array}{r|rr} &e&a&b&c&\\\hline e &e&a&b&c\\ a&a&e&c&b\\ b&b&c&e&a\\ c&c&b&a&e \end{array} But is this a group? If it is, then the table has been realized uniquely. How to check associativity? You don't need to. Consider the map $$ f\colon\{a,b,c,e\}\to S_4 $$ defined by $f(a)=(12)(34)$, $f(b)=(13)(24)$, $f(c)=(14)(23)$, $f(e)$ being the identity. This map respects the operations on the two sets by direct checking and the image is a subgroup of $S_4$.