HyperMath/ Minimum Index Theorem If the index of a subgroup in a group is the smallest prime that divides the order of the group, then the subgroup is normal. BACKGROUND: This is a very useful generalization of the standard result that subgroups of order two are always normal in their containing groups. Theorems of this nature are helpful in analyzing group structure using the Sylow Theorems, for example. KEY DEFINITIONS: 1) index 2) stabilizer 3) conjugate KEY FACT: Subgroup indices are multiplicative, that is if H K G, then G : H G : KK : H. KEY THEOREMS: 1) Representation on Cosets Theorem - we adapt the idea of representing a group element by the permutation it induces via an action on all the elements of a fixed set....in this case a family of mutually conjugate subgroups 2) First Isomorphism Theorem PROOF STRATEGY: Every group induces an action on any of its subgroups simply by multiplication. We will define an action using conjugation of a subgroup with an arbitrary element of the group, then show that for a subgroup satisfying the condition of the theorem (it would be relatively large because its index is small) the stabilizer of the subgroup is actually the entire group. This just means conjugation of the subgroup by any element from the group returns you to the subgroup...and that is normality. PROOF: Suppose H G, G : H p and p is the smallest prime that divides |G|. Define an action : G F F by conjugation, where F be the set of conjugates of H, specifically g,H gHg1. Consider the stabilizer of H: stabH g G : gHg1 H. Certainly g H implies g stabH, so we have stabH H, and hence G : H G : stabH. Now G : H p, and G : stabH divides |G|, so the only possibilities for G : stabH are 1 or p, since there can be no prime factor of |G| strictly smaller than p. Assume for the sake of contradiction that G : stabH p. Since G : H p and stabH H, we conclude stabH H. There are p distinct conjugates of H in G and we can map each element of G into the symmetric group Sp by associating to a given element the specific permutation that it induces among the p conjugates. Let this mapping be denoted by : G Sp. Observe that g1g2 g1 g2, so is a homomorphism. Clearly ker stabH H, since ’ any g ker maps to the identity permutation and therefore conjugation by g doesn t move H. By the First Isomorphism Theorem, Gker im, and in turn im is a subgroup of Sp. It follows that |Gker| divides p! But |Gker| p, because ker H, and G : H p. Therefore it must be exactly p. However since p G : ker G : HH : ker p H : ker, we conclude H : ker 1. This means H ker and then gHg1 H. So G : stabH 1, which is a contradiction. It follows that G : stabH 1. Then G stabH and for any g G, gHg1 H. But then H G. APPLICATION: Show that a group of order 225 32 52 has a normal subgroup of order 75. ’ Sylow s First Theorem gives us at least one 3-SSG of order 9 and one 5-SSG of order 25. N5 1mod5 and N5 divides 225. The only possibility is N5 1, which immediately establishes that the unique subgroup of order 25 is normal. We have at least one 3-SSG with a subgroup of order 3. The internal direct product of this subgroup with the normal subgroup of order 25 is a subgroup of order 75. But the subgroup of order 75 has index 3 in the group of order 225, and by the theorem we have just proved, it must be normal. ’ CAUTION: Just because a prime divides the order of a group, it doesn t mean that there is a subgroup with that index. You have to be sure that a subgroup exists that satisfies the conditions of ’ this theorem. There wouldn t be many simple groups around if it were always true. For example, 2 |A5 | 60 2 3 5. So if there were a subgroup of order 30, it would have index 2 and therefore be normal by the theorem. But A5 is famous for being a simple group....no proper nontrivial normal subgroups. © 2011 Thomas Beatty.