Which of the following multiplication tables defined on the set G = {a, b, c, d}
form a group?
(sorry in advance for the weird formatting of the Cayley tables)
a.
$$◦\ a\ b\ c\ d
\\
a|\ a\ c\ d\ a
\\
b|\ b\ b\ c\ d
\\
c|\ c\ d\ a\ b
\\
d|\ d\ a\ b\ c$$
I think this is not a group:
- associatively fails
- there is no identity element
- thus there is no inverse
$$(a◦b)◦c◦d
\\(c◦c)◦d\neq c◦(c◦d)
\\
a◦d\neq c◦b
\\
a\neq d$$
b.
$$◦\ a\ b\ c\ d\
\\
a|\ a\ b\ c\ d
\\
b|\ b\ a\ d\ c
\\
c|\ c\ d\ a\ b
\\
d|\ d\ c\ b\ a$$
I think this a group:
- associatively property is met $(a◦b)◦c◦d=a◦(b◦c)◦d=a◦b◦(c◦d)$
- there is an identity element (a)
- there is an inverse (each element is an inverse to itself)
$$(a◦b)◦c◦d
\\
(b◦c)◦d=b◦(c◦d)
\\
d◦d=b◦b
\\
a=a$$
$$a◦(b◦c)◦d
\\
(a◦d)◦d=a◦(d◦d)
\\
d◦d=a◦a
\\
a=a$$
$$a◦b◦(c◦d)
\\
(a◦b)◦b=a◦(b◦b)
\\
b◦b=a◦a
\\
a=a$$
c.
$$◦\ a\ b\ c\ d
\\
a|\ a\ b\ c\ d
\\
b|\ b\ c\ d\ a
\\
c|\ c\ d\ a\ b
\\
d|\ d\ a\ b\ c$$
I think this is a group:
- associatively property is met
- there is an identity element (a)
- there is an inverse ($a^{-1}=a$, $b^{-1}=d$, $c^{-1}=c$, $d^{-1}=b$)
$$(a◦b)◦c◦d
\\
(b◦c)◦d=b◦(c◦d)
\\
d◦d=b◦b
\\
c=c$$
$$a◦(b◦c)◦d
\\
(a◦d)◦d=a◦(d◦d)
\\
d◦d=a◦c
\\
c=c$$
$$a◦b◦(c◦d)
\\
(a◦b)◦b=a◦(b◦b)
\\
b◦b=a◦c
\\
c=c$$
d.
$$◦\ a\ b\ c\ d
\\
a|\ a\ b\ c\ d
\\
b|\ b\ a\ c\ d
\\
c|\ c\ b\ a\ d
\\
d|\ d\ d\ b\ c$$
I think this is not a group:
- associatively property is met $(a◦b)◦c◦d=a◦(b◦c)◦d=a◦b◦(c◦d)$
- there is an identity element (a)
- but there is no inverse for d
$$(a◦b)◦c◦d
\\
(b◦c)◦d=b◦(c◦d)
\\
c◦d=b◦d
\\
d=d$$
$$a◦(b◦c)◦d
\\
(a◦c)◦d=a◦(c◦d)
\\
c◦d=a◦d
\\
d=d$$
$$a◦b◦(c◦d)
\\
(a◦b)◦d=a◦(b◦d)
\\
b◦d=a◦d
\\
d=d$$
Do my answers look ok? (specifically unsure about d and whether or not the inverse reason is valid)
Best Answer
It is not a full answer, but two hints, which might help.
In a group, the following implication holds: $ax=ay\Rightarrow x=y$. Therefore on a certain column (or row) each element of the group must appear only once. Therefore, tables a and d are surely not groups.
On the other hand, if you think of known abelian groups of order 4 (from the tables, operations b and c are commutative): you have $\mathbb{Z}_4$ and $\mathbb{Z}_2\times \mathbb{Z}_2 $. If you write the tables for these two groups, you can conclude you have two group isomorphisms.
Hope it helps.