Let N be a normal subgroup of a group G a If G is a cyclic g
Solution
1) Let G be cyclic and N be (normal as G is abelian ) a subgroup of G.
Let x generate G.
Claim: x+N generates G/N
Proof: Any element of G/N is of the form y + N. But y = kx for some integer k (as x generates G)
it follows that y+N = kx + N = k(x+N) .
Hence the result
(2) If G is abelian, then G/N is abelian.
Proof: xN yN = xyN (by definition of quotient group)
=yx N (as G is abelian)
=yNxN
Thus any two cosets commute. So G/N is abelian
(3) If g has order 25 in G , what are the possible orders of gN in G/N?
Now the order of G/N divides the order of G (as the number of elements in G/N is the index [G:N], which divides the order of G.
The order of gN must divide the order of G/N, hence must divide 25.
Possible orders of gN are 1,5,25
