TITLE: Very simple question about symmetric groups: is $\{e\} \subset \ldots \subset S_{n-3} \subset S_{n-2} \subset S_{n-1} \subset S_n$?
QUESTION [5 upvotes]: Is it true to conclude that for any $n \in \mathbb{N}$; $$\{e\} \subset \ldots \subset S_{n-3} \subset S_{n-2} \subset S_{n-1} \subset S_n$$
Intuitively the answer clearly seems yes, since any set of permutations on $n$ letters will also have a subset of all permutations on $n-1$ letters. Does this always hold true?

REPLY [5 votes]: The way I like to see it is that in Algebra, "inclusion" and "injective homomorphism" really mean the same thing. For example, most people would agree that $(\mathbb{Z},+)$ is a subgroup of $(\mathbb{Q},+)$, this latter is a subgroup of $(\mathbb{R},+)$, etc. In this sense, we could say that $\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$ and so on, but formally, what we are really doing is constructing an injective homomorphism for each inclusion. Algebraists usually call them "inclusion homomorphisms". For example, the inclusion homomorphism of $\mathbb{Z}$ in $\mathbb{Q}$ would be:
$$i:\mathbb{Z}\to\mathbb{Q}, n\mapsto\left[\frac{n}{1}\right],$$
since, remember, $\mathbb{Q}$ is formally defined as the quotient field over the ring $\mathbb{Z}$, and therefore, in $\mathbb{Q}$, the number $n$ is really the equivalence class of all fractions which are equivalent to $\frac{n}{1}$.
In the case of this question, as some others have already pointed out, there exists a canonical inclusion homomorphism of $S_{n-1}$ into $S_n$, namely, the one who maps each $\sigma\in S_{n-1}$ to the same permutation in $S_n$, leaving $n$ fixed. Hence, in the algebraic ("up to isomorphism") sense, yes, $S_{n-1}\subset S_n$ just as $\mathbb{Z}\subset\mathbb{Q}$.