For G any group, set T(G) = {g G | o(g) is finite}. If T(G) G, show that T (G /T

Question

For G any group, set T(G) = {g G | o(g) is finite}. If T(G) G, show that T (G /T (G )) = eG/T(G) . That is, if X G /T (G), X T(G)e, then X has infinite order in G /T (G).

Explanation / Answer

1a) If G = , then a ? so that a = a^(kt) for some integer t. ==> a^(kt-1) = 1. Thus, n|(kt - 1) ==> nr = kt - 1 for some integer r. Thus, 1 is a Z-linear combination of k and n ==> gcd(k, n) = 1. Conversely, if gcd(k, n) = 1 we have kr + ns = 1 for some integers r and s. Hence, a = a^(kr + ns) = a^(kr) a^(ns) = a^(kr) * 1 = a^(kr) ? . Therefore, every power of a also is in , and so G = . b) By part (a), a^k is a generator of G gcd(k, n) = 1. The number of such k is given by the Euler phi function. Thus, there are ?(n) such generators. ---------------------------- 2a) If k = 1, this is obvious. If k = -1, then since a^(-1) ? , so is [a^(-1)]^(-1) = a (and vice versa). Thus, = = G. Conversely, suppose that G = . Then (a^k)^r = a^(kr) = a for some integer r. ==> a^(kr-1) = 1. However, a has infinite order. Thus, kr = 1 ==> k (and r) = -1 or 1. b) From part (a), G only has two generators a and a^(-1). ---------------------------- 3) If G is cyclic of prime order, then G has no nontrivial proper subgroups, since every nontrivial element of G is a generator of G (by the first question). Conversely, suppose that G has no nontrivial proper subgroups. Let g ? G different from e. (If there is no such element, then G = {e} and G is cyclic. I suppose this should be included as a cyclic group of prime order in this question.) Consider the set S = {g^n|n = 1, 2, 3, ...} Since this is not {e}, it must be G. Hence G is cyclic with generator g. G can't be of infinite order, because is a proper subgroup of G = . If G is of finite order and not cyclic, then G has a subgroup having order 1

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