Can someone explain what this statement means (Groups and permutations) please?

Question

I'm currently reading notes on a lecture I missed due to not feeling well. These are notes on "Symmetric Groups and Modular Arithmetic Groups".

A sentence in the notes says:

"For a set $S$, a bijection $S\to S$ is called a permutation of $S$."

"Let $G$ be the set of all permutations of $S$ and let $\circ$ be the binary operation of composition, where $a\circ \pi$ means carry out permutation $a$ and then permutation $\pi$. Then $(G,\circ)$ is a group, called the symmetric group on $S$."

Now I understand all of that, apart from where it says:

"where $a\circ \pi$ means carry out permutation $a$ and then permutation $\pi$"

What is this trying to say?

Answer 1

The $\circ$ operation defines the composition, in basically the same way as you can compose functions, except (thanks Mathmo123!) in this definition the composition goes the other way round to what is usual for functions.

For functions (not for permutations!), you are probably familiar with the notation that if $f(x) = x^2$ and $g(x) = x+1$, then $$(g \circ f)(x) = g(f(x)) = x^2+1$$

In this instance (permutations, not functions), using disjoint cycle notation, I might have $a = (1,2,5)$ and $\pi = (3, 5)(2, 1)$. Then $$a \circ \pi = (2,1)(1,2,5)(3,5) = (2,5,3)$$

Answer 2

When you have two bijections $f,g:S \to S$ than you can consider their composition $g \circ f$, just like you can compose functions defined on the real number for example. This composition is also bijective, as the composition of bijections is a bijection.

Usually $g \circ f$ is defined by $ (g \circ f) (s) = g(f(s))$, so you first apply $f$ then $g$.

This is slightly different from what you said but some authors might use the opposite convention. To be sure, I would double check this aspect.