Given a multiplication table for a set $G={a,b,c,d}$, determine whether it is a group.
Given the following multiplication table:
How to determine whether it is a group?
I know that $a$ must be the identity element, since for all $x\in G$, $a \circ x = x \circ a = x$. However, I cannot determine what the inverse element is. Any guess? By the way, the textbook says this is a group, so the inverse element should exist.
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$\begingroup$$a$ is indeed the identity element.
Now observe that $c^2=bd =db =a$ and so $d$ and $b$ are inverse and $c$ is its own inverse. So every element has got its inverse.
Let’s check associativity.
$$(bc)d=d^2=c=b^2=b(cd)\\ (cd)b=b^2=c=ca =c(db)\\ (db)c=ac=c=d^2=d(bc)$$
This is enough because one notices that $bc=cb$, $bd=db $ and $cd=dc$
$\endgroup$ $\begingroup$Well, if you know $a$ is the identity, then look for the positions of $a$ in the table. You'll find it at $(a,a)$, $(b,d)$, $(c,c)$, and $(d,b)$, or in other words, $a^2=a$, $bd=db=a$, and $c^2=a$. So $a^{-1}=a$, $b^{-1}=d$, $c^{-1}=c$, and $d^{-1}=b$.
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