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

• Jul 25th 2011, 07:42 AM
issacnewton
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 $\mathit{e} \in S$ (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 $\mathit{e} \in S$, to show that S is subgroup

of G, we
need to show that $\forall x \in S \;\; x^{-1} \in S$

Since G is finite, S is finite too. Let

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

where $x$ is an arbitrary element of the S. So we need to
show that $x^{-1} \in S$. lets multiply each element of
S by x. So we have set P

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

since S is closed under multiplication , all elements of P are just
rearranged members of S. So $S=P$

By our contruction of S ,

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

But $\mathit{e} \in S$

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

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

since x is arbitrary ,

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

So S is closed with respect to inverses.

So S is a subgroup of G.

Is it correct .... my reasoning ?

thanks
• Jul 25th 2011, 08:04 AM
Swlabr
Re: prove S is subgroup of G where S is stated with some conditions
No - you do not need to prove that $S$ is closed with respect to inverses, you need to prove that $S$ contains inverses - that $x^{-1}\in S$ is $x\in S$. To do this, note that $x$ 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?
• Jul 25th 2011, 09:04 AM
issacnewton
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 $x^{-1} \in S$. is it wrong ?
• Jul 26th 2011, 01:26 AM
Swlabr
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 $x^{-1} \in S$. is it wrong ?

Ah-sorry-your problem is that you need to show that multiplying by $x$ is a bijection, which you haven't done. To do this, it is sufficient to prove that it is an injection. That is, if $xg=xh$ then $g=h$. Can you see how you would do this?
• Jul 26th 2011, 04:49 AM
issacnewton
Re: prove S is subgroup of G where S is stated with some conditions
if $xg=xh \Rightarrow x^{-1}xg=x^{-1}xh \Rightarrow g=h$ , right ? so do I prove injection from S to P ?
• Jul 26th 2011, 04:55 AM
Swlabr
Re: prove S is subgroup of G where S is stated with some conditions
Quote:

Originally Posted by issacnewton
if $xg=xh \Rightarrow x^{-1}xg=x^{-1}xh \Rightarrow g=h$ , 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.
• Jul 26th 2011, 10:03 PM
issacnewton
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 ?
• Jul 27th 2011, 02:23 AM
Swlabr
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?
• Jul 28th 2011, 12:28 AM
issacnewton
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 $\mathbb{N}$
• Jul 28th 2011, 01:50 AM
Swlabr
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 $\mathbb{N}$

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

Now, can you come up with an infinite counter-example?
• Jul 28th 2011, 01:56 AM
issacnewton
Re: prove S is subgroup of G where S is stated with some conditions
counterexample to what ?
• Jul 28th 2011, 01:58 AM
Swlabr
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.
• Jul 29th 2011, 03:09 AM
issacnewton
Re: prove S is subgroup of G where S is stated with some conditions
I think $\mathbb{Z}$ is a group under addition. But $\mathbb{Z}^{+}\cup \{0\}$ would be subset of $\mathbb{Z}$ but the inverses
are not in $\mathbb{Z}^{+}\cup \{0\}$
• Jul 29th 2011, 03:17 AM
Swlabr
Re: prove S is subgroup of G where S is stated with some conditions
Quote:

Originally Posted by issacnewton
I think $\mathbb{Z}$ is a group under addition. But $\mathbb{Z}^{+}\cup \{0\}$ would be subset of $\mathbb{Z}$ but the inverses
are not in $\mathbb{Z}^{+}\cup \{0\}$

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.