prove S is subgroup of G where S is stated with some conditions

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

Re: prove S is subgroup of G where S is stated with some conditions

No - you do not need to prove that $\displaystyle S$ is closed with respect to inverses, you need to prove that $\displaystyle S$ contains inverses - that $\displaystyle x^{-1}\in S$ is $\displaystyle x\in S$. To do this, note that $\displaystyle x$ has finite order. So...

Also, it is interesting to note that this theorem does not hold for infinite groups. Can you come up with a counter-example?

Re: prove S is subgroup of G where S is stated with some conditions

Hi Swlabr

Yes I used wrong wording. But I did prove that $\displaystyle x^{-1} \in S $. is it wrong ?

Re: prove S is subgroup of G where S is stated with some conditions

Quote:

Originally Posted by

**issacnewton** Hi Swlabr

Yes I used wrong wording. But I did prove that $\displaystyle x^{-1} \in S $. is it wrong ?

Ah-sorry-your problem is that you need to show that multiplying by $\displaystyle x$ is a bijection, which you haven't done. To do this, it is sufficient to prove that it is an injection. That is, if $\displaystyle xg=xh$ then $\displaystyle g=h$. Can you see how you would do this?

Re: prove S is subgroup of G where S is stated with some conditions

if $\displaystyle xg=xh \Rightarrow x^{-1}xg=x^{-1}xh \Rightarrow g=h $ , right ? so do I prove injection from S to P ?

Re: prove S is subgroup of G where S is stated with some conditions

Quote:

Originally Posted by

**issacnewton** if $\displaystyle xg=xh \Rightarrow x^{-1}xg=x^{-1}xh \Rightarrow g=h $ , right ? so do I prove injection from S to P ?

Yes, the line you just posted proved that this is an injection. Which completes your proof.

Now, for the sake of completeness, you should work out where you used the fact that the group was finite.

Re: prove S is subgroup of G where S is stated with some conditions

I think when I declared the set S , I only took finite elements. Was that the place I used that fact ?

Re: prove S is subgroup of G where S is stated with some conditions

Quote:

Originally Posted by

**issacnewton** I think when I declared the set S , I only took finite elements. Was that the place I used that fact ?

The finite set bit is used when you are trying to show that there is a bijection between S and P. Can you explain why this would be?

Re: prove S is subgroup of G where S is stated with some conditions

its because , for the set to be finite , there has to be a injection from S to $\displaystyle \mathbb{N}$

Re: prove S is subgroup of G where S is stated with some conditions

Quote:

Originally Posted by

**issacnewton** its because , for the set to be finite , there has to be a injection from S to $\displaystyle \mathbb{N}$

No. You have a map going from $\displaystyle S$ to $\displaystyle P$, $\displaystyle f: S\rightarrow P$. It is surjective, by definition, and you have proven it is injective. Further, $\displaystyle P\subset S$. Now, a finite group has no proper subsets which have the same cardinality, but infinite sets can! Thus, $\displaystyle S=P$ because $\displaystyle S$ is finite.

Now, can you come up with an infinite counter-example?

Re: prove S is subgroup of G where S is stated with some conditions

Re: prove S is subgroup of G where S is stated with some conditions

Quote:

Originally Posted by

**issacnewton** counterexample to what ?

A group which contains a set which is closed under multiplication but not inverses.

Re: prove S is subgroup of G where S is stated with some conditions

I think $\displaystyle \mathbb{Z}$ is a group under addition. But $\displaystyle \mathbb{Z}^{+}\cup \{0\}$ would be subset of $\displaystyle \mathbb{Z}$ but the inverses

are not in $\displaystyle \mathbb{Z}^{+}\cup \{0\}$

Re: prove S is subgroup of G where S is stated with some conditions

Quote:

Originally Posted by

**issacnewton** I think $\displaystyle \mathbb{Z}$ is a group under addition. But $\displaystyle \mathbb{Z}^{+}\cup \{0\}$ would be subset of $\displaystyle \mathbb{Z}$ but the inverses

are not in $\displaystyle \mathbb{Z}^{+}\cup \{0\}$

Yeah, that's a good example. You could also take the integers under multiplication, which is a subset of the rationals under multiplication but obviously not a subgroup.