here is the problem i am trying to do.
let G be a finite group and let S be a nonempty subset of G.
suppose (e is an identity element)
and that S is closed with respect to multiplication.Prove that S
is a subgroup of G.
This is my attempt at solution.
So since , to show that S is subgroup
of G, we
need to show that
Since G is finite, S is finite too. Let
where is an arbitrary element of the S. So we need to
show that . lets multiply each element of
S by x. So we have set P
since S is closed under multiplication , all elements of P are just
rearranged members of S. So
By our contruction of S ,
since x is arbitrary ,
So S is closed with respect to inverses.
So S is a subgroup of G.
Is it correct .... my reasoning ?