The subset of $S_n$ of all permutations that leave some element in its place is a subgroup. Is it a valid example?.

Question

I found this example in appendix A.3 of the book "A terse introduction to linear algebra". But if you permute $\{1,2,3\}$ first keeping only $1$ fixed and then keeping only $3$ fixed, in their composition none of the elements is fixed. So looks like the elements are not closed under the composition operation. Can you point out where am I making a mistake in my reasoning?

Answer 1

I think the book meant the subset leaving a particular element in place is a subgroup, e.g. just the permutations leaving 1 in place is a subgroup. You are correct, the set of permutations leaving any element in place is not a subgroup.

Answer 2

Let $F_1$ be the set of bijective correspondences between $\{1,2,3\}$ and itself such that they fix the element $1$. If $f$ and $g$ are two such permutations, then their composition obviously fixes $1$, by definition of composition of functions, and by the fact that both $f$ and $g$ are in $F_1$. Then observe that the identity function fixes $1$, because it fixes $\{1,2,3\}$. To finally assert that $F_1$ is a subgroup, you need to show that the inverse of a permutation which fixes $1$ still fixes $1$. So suppose $f$ is an element of $F_1$. And let $f^{-1}$ denote its inverse function. Suppose by contradiction that $f^{-1}$ does not fix $1$, then you get the absurd that the composition of $f$ with $f^{-1}$, which is the identity function, does not fix $1$, proving that $f^{-1}$ must be an element of $F_1$. The same for $F_n$ for all $n$, $F_i$ denoting the subset of the set of elements of the symmetric group on $n$ symbols, which fix the $i$-th symbol for every fixed $i$ between $1$ and $n$.

Answer 3

It may worth generalizing a little bit.

Let $A$ be a set and $B\subseteq A$. Define:

$$\operatorname{Fix}(B):=\{f\in \operatorname{Sym}(A)\mid f(b)=b, \forall b\in B\}\subseteq \operatorname{Sym}(A)$$

Let $f,f'\in \operatorname{Fix}(B)$; then, $(ff')(b)=f(f'(b))=f(b)=b, \forall b\in B$, and hence $ff'\in \operatorname{Fix}(B)$. If $\operatorname{Fix}(B)$ is finite (in particular if $A$ is finite), then the closure is enough to conclude that $\operatorname{Fix}(B)\le \operatorname{Sym}(A)$. If $\operatorname{Fix}(B)$ is infinite (then necessarily for infinite $A$), then we have to prove also the "closure by inverses": let $f\in \operatorname{Fix}(B)$; then, $\forall b \in B$, it is $b=f^{-1}(f(b))=f^{-1}(b)$, whence $f^{-1}\in \operatorname{Fix}(B)$.

Yours is the very special case: $A=\{1,2,3\}$ and $B=\{1\}$ $($or $B=\{2\}$, or $B=\{3\}$, or $B=\{1,2\}$, etc.$)$.

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Edit

If we are allowed to appeal to action theory, then by considering the "prototypical" action of $\operatorname{Sym}(A)$ on $A$ (namely, $\sigma\cdot a:=\sigma(a)$), we get:

\begin{alignat}{1} \operatorname{Fix}(B) &=\{f\in \operatorname{Sym}(A)\mid f(b)=b, \forall b\in B\} \\ &=\{f\in \operatorname{Sym}(A)\mid f\in\operatorname{Stab}(b), \forall b\in B\} \\ &=\{f\in\operatorname{Stab}(b), \forall b\in B\} \\ &= \bigcap_{b\in B}\operatorname{Stab}(b) \end{alignat}

Since, $\operatorname{Stab}(a)\le \operatorname{Sym}(A), \forall a\in A$, we have finally $\operatorname{Fix}(B)\le \operatorname{Sym}(A)$.