Let G be a group and {H_i}_I a set of normal subgroups of G so that all G/H_i ar

Question

Let G be a group and {H_i}_I a set of normal subgroups of G so that all G/H_i are Abelian and Intersection H_i =< e_G>.
Show that G is Abelian.

Explanation / Answer

1. i will assume that H is finite (otherwise |H| makes no sense). we need to show that for any x,y in aHa^-1, xy^-1 is also in aHa^-1. so we can write x = ah1a^-1, y = ah2a^-1, for some h1,h2 in H. note that y^-1 = (ah2a^-1)^-1 = (a^-1)^-1(h2)^-1a^-1 = a(h2)^-1a^-1 now xy^-1 = (ah1a^-1)(a(h2)^-1a^-1) = ah1(a^-1a)(h2)^-1a^-1 = a(h1(h2)^-1)a^-1, and since H is a subgroup, h1(h2)^-1 is in H whenever h1 and h2 are. so we can write h1(h2)^-1 = h3, for some h3 in H, and a(h1(h2)^-1)a^-1 = ah3a^-1, which is in aHa^-1. thus aHa^-1 is a subgroup of G. now consider the map f:h-->aha^-1. this is a homomorphism from H to aHa^-1. what is its kernel? suppose f(h) = e, so aha^-1 = e. then ah = a, so h = a^-1a = e. since the kernel is {e}, f must be injective, so H and aHa^-1 have the same order. 2. as we saw in part (1), H and aHa^-1 have the same order, no matter what a is. since H is the ONLY subgroup of order n, and aHa^-1 is a subgroup of order n as well, it must be that H = aHa^-1 for every a in G, which is the definition of a normal subgroup. 3. perhaps you meant when o(a) is finite? suppose that o(a) = k, so a^k = e. (aN)^k = (a^k)N = eN = N, and N is the identity of G/N. thus o(a) must divide o(aN) in G/N. (perhaps it is not clear that if a^n = e, that o(a) = k divides n. to see this, write n as: n = kq + r, where 0 = r

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