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Proof of One Step Subgroup Test

Date: 05/13/2002 at 07:00:04
From: Sarah Hill
Subject: algebra subgroup proof

Hi,

I am stuck with how to do the following proof:

  Prove that a nonempty subset H of a group G is a subgroup 
  of G if and only if a*b^(-1) is in H for all a, b in H.

I am pretty sure that I now have to prove the 4 axioms, that:

  1. G is closed under *
  2. * is associative
  3. the identity e is in G
  4. there exists an inverse element b for every a in G.

I know the operation * is associative as H has the same operation 
as G, but i am unsure how to prove the others, especially the 
inverse and identity ones.

Thank you for your help.

Sarah


Date: 05/13/2002 at 09:06:55
From: Doctor Paul
Subject: Re: algebra subgroup proof

Hi Sarah, 

This is a really powerful theorem.  It's generally known as the 
"One Step Subgroup Test."  Instead of having to show that all of 
the group axioms hold, you only need to show this one result and 
then the group axioms follow.

You wrote:

>I am pretty sure that I now have to prove the 4 axioms, that:
>1. G is closed under *
>2. * is associative
>3. the identity e is in G
>4. there exists an inverse element b for every a in G.

This is for proving the reverse direction -- you're assuming the 
condition holds and then you want to prove that under the 
assumption that H is a subgroup of G.  But the conditions above 
aren't quite right.

We don't want to show (1), (2), and (4) for G.  We are told that 
G is a group so we already know that these hold for G.  We want 
to show them for H.  So we want to show:

  1. H is closed under *
  2. * is associative
  3. the identity e (of the group G) is in H
  4. there exists an inverse element b for every a in H.

The operation * is associative on G (since G is a group) so 
certainly it will be associative on any subset of G.  This shows 
(2).

Now we need to more clearly state our hypothesis:

Suppose that G is a group, H is a nonempty subset of G, and that 
a*b^(-1) is in H for all a, b in H.  We want to use these 
hypotheses to prove (1), (3), and (4).

To see (3), suppose that a*b^(-1) is in H for all a, b in H.

In other words, whenever we pick two elements a and b from H, we 
know that the product a*b^(-1) will be in H.  The key is to make 
clever choices for a and b.  This will become evident below.

Note that we are assuming H is a nonempty subset of G.  So we 
know that H contains at least one element.  Let's consider the 
simplest case -- what if H contained only one element?  

Then if we were choosing elements a and b from H, we would be 
forced to choose a = b.  But we know that whenever we choose two 
elements (call them a and b) from H that a*b^(-1) is in H.  In 
the special case where a = b, we have a*b^(-1) = a*a^(-1) = e is 
in H.  This gives the desired result.  But we have no reason to 
think that assuming H contains only one element is an okay thing 
to do.  So we need to deal with the case where H could contain 
more than one element.

So suppose H is a subset of G that contains more than one 
element.  

We want to pick elements a and b from H.  Here, we are no longer 
forced to pick a = b.  Just pick a = b anyway.  There is nothing 
in the statement of the hypothesis that says we have to pick 
different elements for a and b.  And picking a = b gives the 
desired result (see the previous paragraph).

To see (4) I guess we should consider two cases for completeness. 
We know from above that H contains the identity element.

Case 1:  H has only one element.

We know from (3) that H contains the identity and we are assuming 
H contains only one element so it must be the case that H = {e}.

Certainly e^(-1) = e so for every element in H (there is only one 
element, namely e, for which we need to check this condition)  
there exists another element in H which is the inverse of the 
first element.

Case 2:  H has more than one element.

We know from (3) that H contains the identity element e.  And we 
are assuming it contains at least one other element.  Pick one of 
these additional elements at random and call it x.  We know from 
Case 1 above that there exists an inverse for e in H.  We want to 
show that there exists an inverse for x in H.  To see this, we 
use our hypothesis.  We know that whenever we pick two elements a 
and b from H that a*b^(-1) is in H.  Here, choose a = e, b = x.  
Certainly e and x are in H.  Then e*x^(-1) = x^(-1) is in H.  
This is what we wanted to show.

Now we show (1) -- that H is closed under *.

Suppose x, y are in H.  We can use the hypothesis to note that 
x*y^(-1) is in H.  But that isn't what we want.  We what to 
conclude is that x*y is in H.  Well, if x and y are in H then we 
know by (4) that y^(-1) is in H as well.

Our hypothesis says that whenever we choose two elements a and b 
from H that the product a*b^(-1) will be in H.  Here, choose a = 
x, b = y^(-1).

Then a*b^(-1) = x*y will be in H.

This completes the reverse direction.  Now we need to show the 
forward direction -- that if H is a subgroup then a*b^(-1) is in 
H for all a, b in H.  This is obvious isn't it?  We are assuming 
that H is a group.  So H is closed under taking inverses and 
multiplication.

I believe this completes the proof.  Again, the trick is that you 
have to make intelligent choices about how you choose the 
elements a and b from H.

I hope this helps.  I've tried to make it as clear as possible, 
even including more explanation than I thought was necessary.  I 
did this because I remember having trouble with this proof the 
first time I took a course in modern algebra.  If something 
remains unclear, please write back and we'll work through it.  
Just be sure to let me know where you're having trouble.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Logic
College Modern Algebra

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