Read: Dummit and Foote Chapter 2 Lattice of subgroups of a group (iPad) Subgroups generated by a subset of G Cyclic groups Suppose is a subgroup of G. We will say that is the smallest subgroup of G containing S if S and if H is a subgroup of G containing S then S. Lemma 1. If and 0 both satisfy this denition then = 0. (Proof below that S = S .) h i hh ii Theorem 2. Let G be a group and S a subset. Then there is a unique smallest subgroup of G containing S. First proof (not constructive). Let S = H: h i H a subgroup of G S\H This is the unique smallest subgroup of G containing S. Indeed, these are all subgroups so their intersection is a subgroup: if x; y S then x; y H for all H in the intersection, so 1 2 h i 2 1 xy H and x¡ H and 1 H for all these subgroups, so xy; x¡ ; 1 S proving S is a gro2up. 2 2 2 h i h i Moreover, S is the smallest subgroup of G containing S because if H is any subgroup of G containinhg Si then H is one of the groups in the intersection and so S H. h i Second proof: Let S be dened as the set of all products (words) hh ii a a a 1 2 N 1 where either ai S or ai¡ S and if N = 0 the empty product is interpreted as 1G. I claim this is the smal2lest subgro2up of G containing S. 1 First check that S is a group: if x = a1a2 aN, y = b1 bM S so ai or ai¡ and bi 1 hh ii 2 hh ii or bi¡ are all in S. Then 1 1 1 xy = a a a b b S ; x¡ = a¡ a¡ S 1 2 N 1 M 2 hh ii N 1 2 hh ii and 1G= empty word is in S . Now let us show that thhisii S is the smallest subgroup containing S. Let H be a hh ii 1 subgroup of G contains S. Need to show that a1a2 aN H. Each ai or ai¡ S so ai H so a a a H. This proves that S H. 2 2 2 1 2 N 2 hh ii Now S = S : h i hh ii We proved S is the smallest subgroup of G containing S and S is a subgroup of G containing Sh sio by the property of S we have S S . Thehohthieir inclusion is proved the same way. h i h i hh ii Cyclic groups. If x G we dene the order of x denoted as o(x) or x is smallest positive integer n such that xn2= 1 provided that such an n exists. If xn =/ 1 fojr ajny n > 0 then we say x = . j j 1 A group with a single generator is called cyclic. Proposition 3. Suppose that G = x . This is a group of order (i.e. cardinality) n where n = x . h i j j Proof: I claim that every element of G is one of: 2 n 1 1; x; x ; ; x ¡ and those elements are all distinct. If g G then g = xk for some k because 2 xk k Z f j 2 g is a group containing x and so xk k Z = x . Write k = nq +r with 0 6 r <n. (Division algorithm.) f g f j 2 g h i g = xk = xnq+r = (xn)q xr = xr since xn = 1. It remains to be shown these elements are all distinct. If xa = xb; 0 a < b n 1 6 6 ¡ then b a x ¡ = 1; 0 < b a < n; ¡ contradicting the minimality of n asserted in the denition of x . (Red smallest above.) j j.