Show that a one one map of S onto itself is a permutation group.
You need to remember that a permutation denotes an one-to-one map from S onto itself.
S denotes a set containing a finite number of elements.
A permutation of 4 elements looks like:
`tau = ((1,2,3,4),(tau1,tau2,tau3,tau4))`
If S has n elements (subsets), then the number of permutations is n!.
The law "*" denotes the composition of functions `tau_1` and `tau_2` .
You need to prove all 4 lemmas of a group such that:
Lemma 1: The law "*" on S is closed, hence if `tau_1` and`tau_2 ` `in ` S, then `tau_1*tau_2 ` `in` S.
Lemma 2: The law "*" is associative on S such that`(tau_1*tau_2)*tau_3 = tau_1*(tau_2*tau_3)`
Lemma 3: There is an identity permutation such that `tau = ((1,2,3,4),(1,2,3,4)).`
Lemma 4: The inverse of `tau` maps `tau(1)` into 1.
Hence, if all 4 lemmas hold then (S,*) is a permutation group.