The statement is true.
We know, each element g in G can generate a cyclic group <g>. So if the order of G is 2n+1, G contains at most 2n+1 cyclic subgroups. Now we want to show that G contains at most n+1 ...
In this solution, the concepts of odd order and cyclic groups are investigated. The solution is detailed and well presented.
Group Theory : Homomorphism, Subgroups, Abelian Groups and Group Order
2.Let G be abelian of order n. If gcd(n;m) = 1, prove that f(g) = gm is an automorphism of G. (Note: Automorphism is just an isomorphism from G to itself.)
3. If f : Z7 ! Z5 is a homomorphism, prove that f(x) = 0 for all x 2 Z7.
4. Prove that in the group S10 every permutation of order 20 must be odd.
5. Suppose G is a group in which all subgroups are normal. Suppose further that x; y 2 G are such that gcd(jxj; jyj) = 1.
(a) Prove that x¡1y¡1xy 2 hxi hyi.
(b) Prove that xy = yx.
6. Let H be a subgroup of G, of ¯nite index [G : H] = k. Let C = fgH : g 2 Gg be the set of cosets of H in G.
(a) Prove that for x 2 G the function ¸x(gH) = xgH is a permutation of C.
(b) Let S(C) denote the group of permutations of the set C. Prove that the function F : G 7! S(C) given by F(x) = ¸x is a homomorphism.
(c) Prove that ker(F) · H.
(d) Prove that if jGj = n and n does not divide k!, then ker(F) 6= feg.
(e) Apply this to prove that if jGj = 99 and [G : H] = 9 then H must be a normal subgroup of G.
7. Let G be an abelian group of order 16. Write down all direct products of cyclic groups that could be isomorphic to G. If G has elements x and y, each of order 4, with x2 6= y2, can you determine G (up to isomorphism)?