(a) If p is a prime number bigger than $n$, show that $S_n$ has no element of order $p$.
(b) Decide whether or not $S_7$ has an element of order $12$. Either give an example, or give a proof that no such element exists.
(c) Give an example of an infinite set $S$ and an element $s$ of the group $Sym(S)$ such that $o(s) = 3$.
My attempt:
a) Order of $S_n$ is $n!=n\times (n-1)\times ...\times 2\times 1.$ Since $p$ is prime and greater than $n$, then $\gcd(p,n-i)=1, i = 0,...,n-1.$ Hence, $\gcd(p,n!)= \prod_{i=0}^{n-1}\gcd(p,n-i)=1,$ so $p \nmid n!$ By Lagrange's theorem order of an element must divide order of the group, therefore $S_n$ has no element of order $p$.
b) $|S_7| = 7!,$ and $12\mid 7!$, so by Lagrange's theorem there is an element of order $12$. Example: Consider product of disjoint cycles $(1 2 3)(4 5 6 7)$, then the order is lcm$(3,4) = 12$.
Are there any mistakes in my reasoning? I don't quite understand part(c). Can someone explain about symmetric groups on an infinite set?
If $X$ is any set, the symmetric group $\text{Sym}(X)$ is the set of all bijections $f : X \to X$ where the group operation is composition. The identity element of this group is the identity map on $X$.
So, for example, $f \in \text{Sym}(X)$ has order $3$ if and only if $n=3$ is the smallest natural number having the property that $$f^n(x) = \underbrace{f \circ f \circ \cdots \circ f}_{\text{$n$ times}} (x) \,\,\text{is equal to $x$, for all $x \in X$.} $$ So take your favorite infinite set $X$. Can you write down an order $3$ element of $\text{Sym}(X)$? (Hint: write down some familiar order $3$ elements of finite symmetric groups, express those elements as bijections from a set to itself, and try to find a pattern).