About Finite GroupsDate: 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 (x*y). 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 http://mathforum.org/dr.math/ 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? Thanks, Jason 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 http://mathforum.org/dr.math/ 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. Thanks, Jason 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 http://mathforum.org/library/drmath/view/51852.html or similar pages, which you can find by searching for "cardinality" or "infinite sets" at http://mathforum.org/library/drmath/mathgrepform.html 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 http://mathforum.org/dr.math/ 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! Jason |
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