How should multiplication in groups be interpreted?

Question

Following Cayley, a group $G$ is isomorphic to the image of $$\lambda \colon G\to \operatorname{S}_G,\,a\mapsto \lambda_a$$ where $\lambda_a(x) = ax,\, x\in G$. It is not isomorphic to the image of the analogous map $\rho$ that performs right-multiplication, as the composition of a map $f$ with a map $g$ is written $g\circ f$, not $f\circ g$.

So, is there a "direction" in group multiplication? As it is $ab= \lambda_{ab}(e_G)= (\lambda_a\circ\lambda_b)(e_G)$ but $ab=\rho_{ab}(e_G)\neq (\rho_a\circ\rho_b)(e_G)=ba$, should $ab$ be interpreted as "doing $b$ and, then, $a$"?

I am confused, as lots of visualizations in group-theory use arrows for right-multiplication.

Answer 1

I kind of agree with the sentiment of your question, it's a bit of a pain. But it's possible to make sense of these issues by going in small steps.

To start, group multiplication is a binary operation which inputs an ordered pair $(a,b)$ and outputs a group element (usually written as $ab$ or $a \cdot b$ or $a * b$ or whatever symbol you want to use for your group operation). The order of the elements $a,b$ in an ordered pair certainly matters, and the output of $(b,a)$ (usually written as $ba$ or ...) can be different than the output of $(a,b)$. So far so good.

For some groups, such as $S_G$, the elements are bijective functions from the set to itself, and the group operation is composition. Our convention for function evaluation is almost always to put the function symbol on the left and the argument on the right, obtaining notation like $f(x)$, which is sometimes called "prefix notation" since the function symbol $f$ is the prefix. The group operation in this context is function composition, and the rules of function composition enforce what you have to do: input an ordered pair of group elements $(f,g)$, and output the group element $f \circ g$, whose definition is $$f \circ g(x) = f(g(x)) $$ When it comes to evaluating the expression $f(g(x))$, you should follow the rules you learned in precalculus: start from the innermost parentheses and work your way outward. Doing this, you would evaluate the expression $f(g(x))$ in a few steps:

1. Start with $x$
2. Evaluate $g(x)$
3. Plug the result of that evaluation into $f$
4. Evaluate $f(g(x))$.

So you can see from this that the actual order of evaluations is indeed somewhat backwards: "do $g$, then do $f$". But notice, this has nothing to do with group theory, instead this has to do with prefix notation, our conventions for function evaluation, and the definition of composition.

When you learn the concept of a "group action", you'll see that what I've discussed so far is called a "left group action". There is also a concept of "right group action" (less common than left group action, but not entirely rare) which is associated to suffix notation for functions (very rare, outside of the study of right group actions). See for example the comment of @ancientmathematician

Answer 2

The interpretation "do $b$ first, then $a$" of "$ab$" isn't necessary - we don't need to think of elements of groups being operations - but it is often useful.

Many (most?) groups do in fact arise as the set of functions with some property on a mathematical object - this is called a group action. E.g. the group of permutations of a set, or of linear transformations of a vector space, or of inner automorphisms of another group(!). In these contexts, $ab$ is indeed the element corresponding to the composition of $a$ and $b$.

Now in full generality, there's no interpretation to group multiplication: it's just a binary function on the underlying set of the group, which has certain properties. But intuitively, it may be useful - and it will frequently be appropriate - to think of group elements as functions and of the group operation as composition.

Answer 3

You can actually map it to right multiplication by $a\mapsto\rho_{a^{-1}}$, and this is an isomorphism. There is a degree of symmetry in here, but it depends on the existence of such an antiautomorphism, which not all binary operations have.

Answer 4

Lo mismo aca, puedes visualizarlo así:

$End(G) \times End(G) \rightarrow End(G)$

$(f,g) \mapsto g \circ f$

donde $(f,g)$ sería la operación $(fg)(x) = g(f(x))$