Can one motivate and logically start Group Theory from cosets, as in Lagrange's theorem, instead of the definition of a group?

Question

In Lagrange's theorem one must start with a group $G$ and subgoup $H$.

Let us say instead, sets $H \subset G$ with a binary operation inherited from $G$. Then say we would like the sets to be partionable into distinct cosets under the natural binary coset operation $gH$ where $g \in G$. The fact that they are distinct will force (motivate) an identity, associativity, and inverses via cancellation as follows: If $x$ $\in g_1H$ $\cap g_2H$ $\rightarrow x = g_1h_1 = g_2h_2$ The process of showing $g_1H \subset g_2H$ gives, showing containment in one direction by assuming $g' \in g_1H$ or $g' = g_1h'$ for some $h' \in H$

The point here: we must use associativity inverses and identity to solve and get $g_1=g_2h_2h_1^{-1}$ in a natural or normal way. These are exactly what we need for the definition of a group! This is why I think Group Theory could start with Lagrange's theorem as a motivator for the definition of a group. We could say that "If this is to work then the group properties are exactly what we need."

Can you show that this fails to force the definition of a group in any way? Is this logically adequate to create the definition of a group starting with cosets (as equivalence classes)? Can you get to distinct cosets without forcing the cancellation process to work? Please explain.

Answer 1

You are incorrect in thinking that associativity and inverses are required for the cosets to form a partition.

Consider the three element quasigroup that is not a group, $$\begin{array}{c|ccc} \cdot & e & a & b\\ \hline e & e & a & b\\ a & b & e & a\\ b & a & b & e \end{array}.$$ The operation is not associative: $(ae)b = bb = e$, but $a(eb)=ab=a$. The only nonempty subsets that are closed under the operation are $\{e\}$ and $\{e,a,b\}$, since having either $a$ or $b$ will give you $e$, and with $e$ and any other element you get all of them.

The "cosets" of $\{e,a,b\}$ are just $\{e,a,b\}$, of course. For $\{e\}$, we get exactly $\{e\}$, $\{a\}$, and $\{b\}$, which are disjoint.

Note that we have no identity element, so we can't have inverses. We do have cancellation, because the equations $rx=s$ and $yr=s$ have unique solutions for all $r,s\in\{e,a,b\}$.

So just asking for a nonempty set with a binary operation such that any nonempty subset that is closed under the operation will partition the set (even requiring that the "parts" have the same size) does not force you to get a group. Associativity is not forced, and inverses are not required for cancellation to hold (cancellation holds in the positive integers under addition, but there are no inverses).

I suspect the "natural or normal way" that you perceive is born out of expectation and familiarity, not out of any actual "naturalness" or "normality". Kind of like the water marveling at how precisely the pothole it resides in is shaped to accommodate it.

Answer 2

The basis elements of the octonions together with their negatives form a loop (quasigroup with identity element) of order 16 which is called the octonion loop (or Cayley loop). This is analogous to how the basis elements of the quaternions together with their negatives form the quaternion group.

The octonion loop is an example of a nonassociative loop in which every subloop is normal. This implies that the cosets of each subloop partition the whole loop. So no, associativity is not necessary to have coset partitions.