ABSTRACT ALGEBRA I: HOMEWORK 7 SOLUTIONS ANDY EISENBERG Problem (Section 3.6, Problem 18). Let S be an infinite set. Let H be the set of all elements σ ∈ Sym(S) such that σ(x)= x for all but finitely many x. Prove that H is a subgroup of Sym(S). Proof. Let σ, τ ∈ H. Let A ⊂ S be the set of elements not fixed by σ, and B ⊂ S the set of elements not fixed by τ. By assumption, A and B are finite sets, so A∪B is a finite set. Let x∈ / A ∪ B. Then σ(τ(x)) = σ(x) = x, so στ leaves all but at most finitely elements of S fixed, i.e. στ ∈ H. Now observe that if y∈ / A, then σ(y) = y, so σ−1(y) = y. Thus σ−1 ∈ H, and H is a subgroup. Problem (Section 3.6, Problem 19). The center of a group is the set of all elements that commute with every other element of the group. That is, Z(G) = {x ∈ G | xg = gx, ∀ g ∈ G}. Show that if n ≥ 3, then the center of Sn is trivial. Proof. Let σ ∈ Sn. If σ 6= e, then σ(i) = j for some i 6= j. Let k 6= i, j (such a k must exist since n ≥ 3), and let τ = (i k). Then σ(τ(i)) = σ(k) 6= j = τ(j)= τ(σ(i)). Since for every non-identity element of Sn we have found an element which does not commute with it, the center of Sn cannot contain any non-identity element. Since the center always contains the identity element, we have Z(Sn)= {e}. Problem (Section 3.6, Problem 20). Let the dihedral group Dn be given by ele- ments a of order n and b of order 2, where ba = a−1b. Find the smallest subgroup 2 of Dn that contains a and b. Hint: Consider two cases, depending on whether n is odd or even. 2 n+1 +1 Proof. Let n be odd. Then (a ) 2 = an = a. By definition, {a,b} is a generating 2 set of Dn. Since the subgroup generated by a and b contains a and b, it contains all of Dn. Now let n be even. Then a2 has order n/2, b has order 2, and ba2 = a−1ba = a−2b = (a2)−1b. Since these properties define the dihedral group, we have ha2,bi is isomorphic to Dn/2. Problem (Section 3.6, Problem 21). Find the center of the dihedral group Dn. Hint: Consider two cases, depending on whether n is odd or even. k −k k k Proof. Let n be odd. Observe that ba = a b 6= a b, so b∈ / Z(Dn) and a ∈/ k Z(Dn). Similarly, ba ∈/ Z(Dn), since − − − abak = aa kb = a1 kb = bak 1 6= bak+1 = baka. Thus Z(Dn) =∼ {e}. Date: Fall 2009. 1 2 ANDY EISENBERG If n is even, then a−kb = akb if and only if k = n/2, so by the same argument as k k before, b∈ / Z(Dn) and a ∈/ Z(Dn) for k 6= n/2. As above, ba ∈/ Z(Dn). Observe that − − − bai(an/2)= bai+n/2 = a i n/2b = an/2a ib = an/2bai, so an/2 ∈ Z(G). Thus Z(G) =∼ {e,an/2}. .