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

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?

Answer 1

It's interesting that you appear to be getting contradictory answers to this apparently simple question. In fact the confusion is resulting from the fact that we don't usually bother to define $S_n$ precisely, and this makes the question difficult to answer definitively.

$S_n$ tends to be used to denote an arbitrary member of an isomorphism class of groups. With that convention, specialists in the area would generally agree that $S_{n-1}$ is a subgroup of $S_n$, but that is a little informal and it relies on the fact that your audience will know exactly what you mean.

Personally, I prefer to use ${\rm Sym}(X)$ to denote the group of all permutations of a set $X$ (that's totally specific), I use ${\rm Sym}_n$ or ${\rm Sym}(n)$ to denote the group of permutations of an arbitrary set $X$ with $|X|=n$ (which is less specific, but you can always allow $X$ to default to $\{1,2,\ldots,n\}$), and $S_n$ to denote an arbirary group isomorphic to $S{\rm Sym}(n)$.

PS: Let's see what GAP thinks about it:

gap> IsSubgroup(SymmetricGroup(6),SymmetricGroup(5));
true

Now we know! In fact GAP has the same viewpoint as in the answer of user1729, it regards all permutations as living in ${\rm Sym}({\mathbb N})$. That can sometimes be incovenient, because it means that permutation groups do not have a well-defined degree - they just a largest moved point.

Magma disagrees however:

> Sym(5) subset Sym(6);
false

Answer 2

I think that it is false because if you take an element $\sigma$ of $S_n$ it will not be defined for $n+1$ therefore $\sigma \not \in S_{n+1}$.

Edit : could you please explain why was I downvoted twice? Did I say something wrong?

Answer 3

The answer is "yes" if you think of $S_n$ as the set of permutations of the particular $n$-element set $\{1,2,\ldots,n\}$. Then $S_{n-1}$ is (essentially) the set of permutations that fix $n$. That is a natural convention, but not a mathematical property of the group.

In general, there are $n$ ways to choose an $n-1$-element subset of an $n$ element set. Each of those defines an injection of $S_{n-1}$ into $S_n$.

Answer 4

No.

It is true that, for each $n\in \Bbb N$, we have some subgroup $G$ of $S_{n+1}$ for which $S_n\cong G$.

Each element of (the underlying set of) $S_n$ is a bijection from $N=\{1,\dots, n\}$ to $N$. This bijection can be extended in the obvious way to a bijection from $M=\{1,\dots, n, n+1\}$ to $M$, but the original element is not equal to such an extension.

Answer 5

Let me add yet another angle: We often view elements of the $S_n$s using disjoint cycle notation, so for example $(12)(345)$ is an element of $S_5$. However, $(12)(345)$ also defines an element of $S_6$, and in fact of $S_n$ for every $n\geq5$.

Therefore, in this view we are really thinking of $S_n$ as a subgroup of $\operatorname{Sym}(\mathbb{N})$, and so here, in this super-common setting, we do get the claimed chain: $$\{e\} \subset \ldots \subset S_{n-3} \subset S_{n-2} \subset S_{n-1} \subset S_n$$

I should stress though that this is about an extremely specific view of these groups, and is not about the abstract groups themselves.

Answer 6

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 $\mathbb{Z}\subset\mathbb{Q}$.