Set acting like a Group

Question

I am a little confused with familiar things, so I am looking for some help.

Description:

Consider the set, $S_3= \{(123), (132), (213), (231), (321), (312)\}$, a symmetric group acting on $3$ objects. Let us consider another set $A=\{a,b,c,d,e,f\}$. Consider relation $R : S_3 \rightarrow A$ where $R$ relates each element of $S_3$ to only one element of $A$.This analogous to a bijection . This relation can be done $6!=|A|!$ ways, $R$ is one of the way (please, let me know if this is not clear).

Question:

1. Can I say $A$ forms a group?

   My point is that we can write each element of $A$ as a product of elements of $A$ due to relation $R$. For example, if $\pi_1, \pi_2 \in S_3$ is related to $a,b \in A$, then we can say , there exists $ \pi_3 =\pi_ \circ \pi_2 \in S_3$. Due to $R$, $\exists x \in A$ where $x$ is related to $\pi_3$, so, $a \circ b =x$, thus $A$ has a concept similar to closure.

2. Is there similar concept/terminology?

Answer 1

If I understand you correctly, the relation $R$ you describe is not just like a bijection, it is a bijection. Specifically, $(A,R,B)$ with $R\subseteq A \times B$ is a function if: $$\forall a\in A : \exists! b\in B : (a,b)\in R$$ and such a function is called a bijection, if:

$$\forall b\in B : \exists! a\in A : (a,b)\in R$$

It is not surprising, that $A$ "forms a group". You can "transport" all and every kind of structure on a set along any bijection into another set. You find nothing new, however: $A$ together with the operation you describe is still $S_3$ (up to isomorphism, but that is what matters). The situation is of course the same for other kinds of structures on your set.

Answer 2

From a purely pedantic formal perspective: No, merely defining the bijection between the permutation set and A does not define a group; you need to also provide the group operation as a relation on A. Of course, I understand that what you intend is implicitly obvious, but implicitness doesn't count in definitions. The point is that, formally, you don't have a group definition until you explicitly write down the relation - or more usefully, the rule used to build the relation given the permutation set and the bijection.

Of course, you could just add the phrase 'with the obvious induced operation', or something like it, at the appropriate point, and mathematicians would typically accept that as discharging the obligation.