# Odd Order and Cyclic Groups

Suppose that G is a finite group of odd order 2n + 1. Prove or disprove that the number of cyclic subgroups of G is at most n + 1.

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#### Solution Preview

The statement is true.

Proof:

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 ...

#### Solution Summary

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

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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.

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