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.

... the group S10 every permutation of order 20 must be odd. ... 7. Let G be an abelian group of order 16 ... Write down all direct products of cyclic groups that could be ...

...cyclic permutations over 2 elements will be the odd ones, (23 ... A Sylow p-subgroup of order p will of course be ... SY1 will Z120 have Sylow p-subgroups of orders 4=2 ...

... What is the order of S4 X S4 ... A finite group G has elements of orders p and q, where p and q are ... since, for any integer x, the number 2x − 1 is odd, so cannot ...

... as the order of x2 then the order of x is odd. Find the order of each of the following elements in the given ... Deﬁne what it means for a group to be cyclic. ...

... nZ has a unique (cyclic) subgroup of order d for ... classes of are if n is odd and if ... three problems involving quaternions as well as dihedral and cyclic groups. ...

... p^2. Given that fact, show that for any odd prime p ... there is a number b such that b has order p(p ... p^2. (Note that b is the generator of the cyclic group of units ...

... permutations of objects that are not all different, and cyclic notation. ... 3. You can order one item from a list of 5 ... 3. How many odd 2-digit numbers are there? ...

... substituting 1 will give 0 only if there are an odd number of x ... dividing the discriminant of f , the decomposition group is a cyclic group of order at most ...

... as the order of x2 then the order of x is odd. Find the order of each of the following elements in the given ... Deﬁne what it means for a group to be cyclic. ...

Let G be a finite group of order 2k, k odd, that contains a cyclic group of oder k. Determine a formula to compute the number of subgroups of G that are of odd...