Homework Sourse1 Let G be a group of order 2 if n2 prove that G has no subg
1) Let G be a group of order 2
if n>2, prove that G has no subgroup of order n-1.
2) Let G be a group. Given x in G and K subgroup G, define
C-G(x)={ g in G: xg=gx}
C_G(K) ={ g in G : gk=kg for all k in K}
a) Prove Z(G) is subgroup of C_G(a) fro every a in G.
b) Prove that G is Abelian if and only if G=C_G(a) for every a in G
Solution
Solution Let G be a group of order 2m where m is odd. Prove that G contains a normal subgroup of order m
Let |G|=2nm|G|=2nm where 2m2m. If G has a cyclic 2- Sylow subgroup, then G has a normal subgroup of order m.
When n=0, the result is clear. Now consider the case when n1. First, there is a normal subgroup of GGof order 2n1m 2n1m. In order to proof this, consider the permutation of G on itself by left multiplication, :GSymG As G has a cyclic 2-Sylow subgroup, G has an element g of order 2n, and (g) is the composition of m 2n-cycles. Thus,(g) is an odd permutation. So, the composition map gn:G{±1}is onto. The kernel of this map is a normal subgroupH of G of order 2n1m
When n=0 the result is clear. Now consider the case when n1. First, there is a normal subgroup of Gof order 2n1m. In order to proof this, consider the permutation of G on itself by left multiplication, :G:GSymG As G has a cyclic 2-Sylow subgroup, G has an element g of order 2n, and (g) is the composition of m 2n-cycles. Thus, (g) is an odd permutation. So, the composition map sgn:G{±1} is onto. The kernel of this map is a normal subgroup H of G, of order 2n1m
The two are of the same order, thus equal. Now, for any xG xNx1xHx1=HxNx1xHx1=H, so xNx1x is another subgroup of H of order m. So, xNx1=N. is normal in G of order m
solution for B
f G/Z(G) is cyclic then G is abelian.
f G/Z(G) is cyclic with generator xZ(G). Every element in G/Z(G) can be written as x k z for some k Z and z Z(G). Now, let g, h G, then g = x a z and h = x bw for z, w Z(G). We have gh = x a zxbw = x a+b zw = x b+awz = x bwxa z = hg.
