202004240606 Homework 1 (Q1)

Let $G=\{ a,b,c,d\}$ and the Cayley table of $G$ be

$\begin{aligned} * \quad | &\textrm{} \quad a &\textrm{}\quad b \quad &\textrm{}\quad c &\textrm{}\quad d \\ \hline a \quad | &\quad a &\quad b \quad &\quad c &\quad d \\ b \quad | &\quad b &\quad a \quad &\quad c &\quad d \\ c \quad | &\quad c &\quad b \quad &\quad a &\quad d \\ d \quad | &\quad d &\quad d \quad &\quad b &\quad c \end{aligned}$

Is $(G,*)$ a group?

Attempts.

I recall that a group $(G,*)$ is a set $G$ altogether with an operation $*:G\times G\rightarrow G$ , $(x,y)\mapsto x*y$ such that the following axioms be satisfied:

i. (Associativity) $(x*y)*z=x*(y*z)$ ;

ii. (Identity) There exists $e\in G$ such that $e*x=x*e=x$ for all $x\in G$ ; and

iii. (Inverse) For each $a\in G$ , there exists $b\in G$ such that $a*b=b*a=e$ .

Let me check them one by one.

i. For $x,y,z\in \{ a,b,c,d \}$ , it suffices to check $4\times 4\times 4=64$ operations. And I found that the following operation failed axiom (i):

$d*(c*b)=d*b=d$ but $(d*c)*b=b*b=a$ .

That said, $G$ under the operation $*$ is not a group.

ii. By inspection, the identity of $G$ exists, and that is $a$ because $a*x=x*a=x$ for all $x\in \{ a,b,c,d\}$ . Axiom (ii) is thus satisfied.

iii. There exists $d\in G$ , such that for all $x^i\in\{ a,b,c,d\}=G$ ,

$d*x^i=x^i*d\neq a$ .

The claim above is validated by simply checking the identity found to be $a$ in part ii. cannot be found in the entries, and by the fact that $a$ is unique.

Axioms i. and iii. having been failed, I may conclude that $G$ under the operation $*$ cannot form a group.

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