Show that every cyclic group Cn of order n is abelian. (Moreover, show that if G is a group, so is GxG)

Solution Preview

1. Cn is a cyclic group of order n. Then Cn=<a>, where a is a generator of Cn. So for any elements x,y in Cn, we have x=a^i, y=a^j for some integer i,j. Thus ...

Solution Summary

It is shown that every cyclic group Cn of order n is abelian. The proof is concise.

Let G be a nonabelian group and Z(G) be its center. Show that the factor group G/Z(G) is not a cyclic group.
We know if G is abelian, Z(G)=G. But now if it is not abelian, can we simply say because G is not cyclic, then any factor group will not be cyclic either? or is there more to it?

Determine all binery cyclic codes of length 5.
Note: To find all cyclic codes of length n, find all ideals in B[x]/x^2+1
Note: If 1 is an Ideal (I) then R = I.
Example:
n=2
R=B[x]/x^2+1, x^2=1
R={o,1,x,1+x}
Ideals <0> = 0
<1> = R
x = (0, x, x^2...)
= (1,...

Thisquestion is concerned with subgroups ofthe group S5 of permutations on the set {1,2,3,4,5} , a group with 120 elements.
(a) Explain why this group has cyclic subgroups of order 1,2,3,4,5 and 6, and give examples of each of these.
Explain why this group does not have cyclic subgroups of any other order.
(8 marks
(b) By co

Note: ~~ means an isomorphism exists. Moreover,if an isomorphism existed from G to G1 I would say G ~~ G1
Questions: If G is an infinite cyclic group, show that G ~~ Z (Z is the set of integers)

Problem 1.
Let a,b be elements of a group G
Show a) the conjugate of the product of a and b is the product of the conjugate of a and the conjugate of b
b) show that the conjugate of a^-1 is the inverse of the conjugate of a
c)let N=(S) for some subset S of G. Prove that the N is a normal subgroup of G if
gSg^-1<=N for

? I have the following cayley tables (which is in modulo 9)
determine the order of each element . Prove that G is a cyclic group.
? Let be the symmetric group of degree 3 together with composition of maps. Is G isomorphic to ? Justify your answer.
? Let p be a prime number and G a group of order with identity e

Please see the attached file first.
I did the proof, but it's weak, since I can't find a way to argue why every odd order subgroup has to be a subgroup of the cyclic odd order subgroup K of index 2, and by the divisibility argument it still could be a subgroup of G and not to be contained entirely in K.