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Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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UNIT-II<br />

61<br />

of nilpotency of G) , so Z c<br />

(G) = G. Now G'(<br />

. So we get the derived series<br />

.<br />

Hence G is solvable.<br />

Note that the converse is not true i.e. a solvable group need not be nilpotent. For example :<br />

G ∆A<br />

∆bg<br />

3 3 1<br />

S 3<br />

is solvable but is not nilpolent because center of S 3<br />

is (1). (Every nilpotent group has a non-trivial center)<br />

Exmple 13 :<br />

Consider<br />

b g bg b g bg<br />

. Now Z Z × S = Z × 1 and every Z × S = Z × 1 . Therefore, ascending<br />

1 2 3 2<br />

Z k 2 3 2<br />

central series (or upper central series) never reach G. Hence Z<br />

× S is not niloptent.<br />

2 3<br />

( )<br />

l q l1q<br />

c<br />

Zc−G1 , GG' cZe<br />

2 , − G,<br />

c −1<br />

G = ∆ Z b g (<br />

' ∆ ' ∆ ∆ ∆=<br />

1<br />

c )<br />

2<br />

× S3<br />

− − − G =

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