Subgroup and Order of a GroupDate: 11/06/2003 at 11:41:57 From: Lori Subject: Modern Algebra (Subgroup And Order of a Group) If o(G) = p^n, p is a prime number, and H is not equal to G and it is a subgroup of G, show that there exists an x that is an element of G and x is not an element of H such that x^(-1)Hx = H. I find it confusing to figure out x^(-1)Hx = H for an element x. If H is normal in G, then clearly x^(-1)Hx = H for all x that are elements of G. But then if H is just a subgroup of G, then x^(-1)Hx won't necessarily equal H unless proven. I figure that o(G) = p^n is the reason for this to be true, but don't know if Sylow's theorem or something else needs to be used. Thanks, Dr. Math has helped me figure out how to start a couple of problems before and was very helpful. Date: 11/07/2003 at 03:40:39 From: Doctor Jacques Subject: Re: Modern Algebra (Subgroup And Order of a Group) Hi Lori, We need two results to prove this: 1) In a group of order p^n, where p is prime, the center of the group is not trivial. This is proven by counting elements in conjugacy classes. 2) The third isomorphism theorem: if f : G -> L is a homomorphism of groups, f defines a bijection between the subgroups of f(G) and the subgroups of G containing ker f. A consequence of this is that if H1 and H2 are two subgroups of G containing ker f, and f(H1) = f(H2), then H1 = H2. If you are not familiar with either of these facts, please write back. Let G be a group of order p^n, p prime, and H be a proper subgroup of G. Let C be the center of G. If C is not a subgroup of H, this means that C contains an element x, not in H, that commutes with all elements of G, and, in particular, with all elements of H. In that case, we have x^(-1)*h*x = h for all h in H, and we are done. We may therefore assume that C is a subgroup of H. We proceed by induction on n. If n = 1, |G| = p, and the only proper subgroup of G is {e}--the result is trivial in this case. Let us assume, by induction, that the theorem is true for all groups of order p^k, for k < n. As C is normal in G, we may consider the canonical homomorphism f : G -> G/C = L. As C is not trivial, L is of order p^k with k < n, and the theorem holds in L by the induction hypothesis. This means that there is an element b in L, not in f(H) such that: b^(-1)*f(H)*b = f(H) (Note that f(H) is a proper subgroup of L, because otherwise we would have f(H) = f(G) and H = G by property (2) above.) Now, b = f(a) for some a in G, and we have, by the homomorphism property: f(a^(-1)*H*a) = f(H) We can remark: * a is not in H, because f(a) = b is not in f(H) * C is a subset of a^(-1)*H*a, because, for any c in C, we have c = a^(-1)*c*a, and the latter is in a^(-1)*H*a, as C is a subset of H by hypothesis. Can you conclude the proof? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
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