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About Finite Groups

Date: 02/03/2003 at 00:18:27
From: Jason
Subject: Question about finite groups

If H is a nonempty subset of the finite group {G,*} with the property 
that x*y is in H when x and y are in H, is H a subgroup of G? How is 
the finiteness used?

Obviously the only thing left to do is prove that H contains the
inverse of each of its elements. I can't seem to do that.

Date: 02/03/2003 at 03:02:29
From: Doctor Jacques
Subject: Re: Question about finite groups

Hi Jason,

Note that you must also prove that H contains the identity.

Now, if G is finite, so is H. Let n be the number of elements in H.

Consider, for a fixed x in H and all y in H, the set of elements 

We know by hypothesis that all these elements lie in H.

Assume that two of these elements are equal :

  x*y_1 = x*y_2

Since G is a group, x has an inverse in G (we do not know yet that 
this inverse is in H). However, we are still able to multiply by
x^(-1) in G and conclude that y_1 = y_2.

This means that the n elements x*y are distinct and lie in H (a set 
of n elements). We can therefore conclude that these elements are all 
elements of H in some order.

In particular, since x is in H, there is a y in H such that x*y = x.

This means that y is the identify (e) and is in H.

Now, you should be able to use the same kind of reasoning to show 
that x^(-1) is in H.

If you are still stuck, please feel free to write back

- Doctor Jacques, The Math Forum 

Date: 02/03/2003 at 17:10:45
From: Jason
Subject: Question about finite groups

So, if I say that for fixed x and x*y there is a y such that x*y
= e, and therefore y=x'have I proved an inverse for EVERY element x?


Date: 02/04/2003 at 01:58:47
From: Doctor Jacques
Subject: Re: Question about finite groups

Yes, that is correct. Congratulations.

- Doctor Jacques, The Math Forum 

Date: 02/05/2003 at 02:38:13
From: Jason
Subject: Re: Question about finite groups

What if the subset of G is not finite but the other conditions hold?  
Can we still use the same process? (i.e.multiply by a fixed x to get a 
distinct permutation?), then from there claim that there is a y such
that x*y=x?  

If so, then the finiteness is not necessary. So, my real question is 
why is finiteness necessary for the subset to be a subgroup?

Hope that made sense.

Date: 02/05/2003 at 02:54:39
From: Doctor Jacques
Subject: Re: Question about finite groups

Hi Jason, and thank you for writing again,

The finiteness is used in this way:

We show that we can find a set S of n elements, all distinct, in a 
set T of n elements. This implies that all the elements of S are in 
fact the elements of T in a possibly different order.

To state it differently, if a finite set T of n elements contains a 
subset S of n elements, then T = S.

This is not true for infinite sets. Of course, we must first define 
what is meant by "the same number of elements" for such sets. You 
should have a look at:

   Infinite Sets 

or similar pages, which you can find by searching for "cardinality" 
or "infinite sets" at 

For example, the set Z of integers and the set 2Z of even integers can 
be considered equivalent (there is a bijective correspondence between 
them), but they are not equal.

In our case about groups, there is a counterexample:

Consider the multiplicative group of non-zero rationals. In that 
group, the set of non-zero integers satisifies the property you 
mention; namely for any x, y non-sero integers, x*y is also an non-
zero integer. However, this is not a subgroup, since most elements 
(like 2) do not have a (multiplicative) inverse in the subset.

I hope this clarifies things.

- Doctor Jacques, The Math Forum 

Date: 02/05/2003 at 16:38:28
From: Jason
Subject: Thank you (Question about finite groups)

Thanks so much for the help! You guys are great!
Associated Topics:
College Modern Algebra

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