This is not exactly a question but I want someone to confirm if my understanding of the following is correct.
The symmetric group of a set $X$ is the set of all the bijections from $X \to X$. $$S_{X}= \{\sigma: X \to X \mid \sigma \text{ is a bijection} \}$$
Question 1: I want to show the above set is a group with respect to the binary operation of composition $\circ : S_X \times S_X \to S_X$. Here is my proof.
- Since $\sigma_1, \sigma_2 \in S_X$ are bijections, their composition is also a bijection and hence belongs to $S_X$.
- Composition is associative for any triplet of functions (not necessarily bijections) so associativity holds.
- $\sigma_e : X \to X$ given by $\sigma_e (x) = x$ is the identity element in $S_X$.
- Any $\sigma \in S_X$ is bijective, hence invertible and so $\exists \ \sigma^{-1} \in S_X$ such that $\sigma \circ \sigma^{-1}= \sigma_e$
Something about point 3 and 4 doesn't feel right. What is a better way to phrase it?
Is the reasoning in my proof above correct at every step?
Question 2: I'm confused about Symmetric group vs Permutation Group. But as I understand it, a symmetric group is the set of all the bijections and permutation group is just a subgroup of symmetric group including the Symmetric group. Intuitively, a permutation group is just a collection of bijections on $X$ that form a group with $\circ$. So, one could say: There is only one symmetric group on a set while there could be multiple permutation groups on the same set. Is my understanding correct?
For example: $P = \{\pi : ℝ \to ℝ \mid \pi(x) = x+a, \ \forall a \in ℝ \} $ is set of all translations (as well as bijections) is the Permutation group on $\mathbb{R}$ but $P≠S_{ ℝ}$. $P$ is just the group $(ℝ, +)$.
