Hi

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 ,

But

So

since x is arbitrary ,

So S is closed with respect to inverses.

So S is a subgroup of G.

Is it correct .... my reasoning ?

thanks