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 $\displaystyle \mathit{e} \in S$ (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 $\displaystyle \mathit{e} \in S$, to show that S is subgroup

of G, we

need to show that $\displaystyle \forall x \in S \;\; x^{-1} \in S$

Since G is finite, S is finite too. Let

$\displaystyle S=\{\mathit{e},a_1,a_2 \cdots ,x,\cdots a_n \}$

where $\displaystyle x$ is an arbitrary element of the S. So we need to

show that $\displaystyle x^{-1} \in S$. lets multiply each element of

S by x. So we have set P

$\displaystyle P=\{x\ast \mathit{e},x\ast a_1 \cdots ,x \ast x \cdots ,x \ast a_n \}$

since S is closed under multiplication , all elements of P are just

rearranged members of S. So $\displaystyle S=P$

By our contruction of S ,

$\displaystyle x\ne \mathit{e} \;\Rightarrow \;x\ast \mathit{e} \ne \mathit{e}$

But $\displaystyle \mathit{e} \in S$

So $\displaystyle \exists y \in S \;\; (x\ast y =\mathit{e})$

$\displaystyle \Rightarrow y = x^{-1} $

since x is arbitrary ,

$\displaystyle \forall x \in S \;\; x^{-1} \in S$

So S is closed with respect to inverses.

So S is a subgroup of G.

Is it correct .... my reasoning ?

thanks