Normal Subgroups

James F. Hurley

Copyright 2001 by James F. Hurley, Department of Mathematics, University of Connecticut, Unit 3009, Storrs CT 06269-3009. All rights reserved.

Maple's group package can determine whether a subgroup H of a group G is normal. After invoking the group package, you need to specify both G and H . You then can ask Maple whether H "i snormal " in G .

As in the
Groups and Cosets worksheet, the second command below defines the symmetric group S[3] of degree 3 as the group generated by the transposition g = ( 2 3 ) and the 3-cycle f = ( 1 2 3 ). The next command defines the subgroup H to be the cyclic group generated by f . Since we know that H is a subgroup of order 3 in a group of order 6, it has index 2 and so is indeed a normal subgroup. The last line shows the syntax for the isnormal command. Execute the routine to see Maple's answer to the question "is H normal in G "? [Recall that to execute the commands, position the cursor at the end after the last semicolon and hit the Enter key (or press the Shift and Return keys simultaneously).]

> with(group):
S_3 := permgroup( 3, {[[2, 3]], [[1, 2, 3]]} );
H := permgroup( 3, {[[1, 2, 3]]} );
isnormal(S_3, H);

S_3 := permgroup(3,{[[1, 2, 3]], [[2, 3]]})

H := permgroup(3,{[[1, 2, 3]]})

true


Could it be that the cyclic subgroup H generated by the 3-cycle ( 1 2 3 ) is always normal in S[n] , no matter the value of n ? To investigate, let's ask Maple whether the subgroup generated in S[4] by that 3-cycle ( 1 2 3) is normal. Note that S[4] is generated by all transpositions ( i j ), of which there are exactly 6 = (4*3)/2.

> G := permgroup( 4, {[[1, 2]], [[1, 3]], [[1, 4]], [[2, 3]], [[2, 4]], [[3, 4]]} );
H := permgroup( 4, {[[1, 2, 3]]} );
isnormal(G, H);

G := permgroup(4,{[[2, 3]], [[1, 2]], [[1, 4]], [[1...

H := permgroup(4,{[[1, 2, 3]]})

false


Maple's answer kills the conjecture about
H being normal in S[n] for every n !

The
normalizer N ( H ) of a subgroup H of a group G is the largest subgroup of G in which the subgroup H is normal (see Exercise 11.23, p. 104, of D. Saracino, Abstract Algebra: A First Course , Waveland Press, 1992.) Maple has a command to determine the normalizer of a given subgroup of a given group G . The following calculation calculates the normalizer of the non-normal subgroup H of S[4] . Notice that the syntax requires specification first of the group G inside which the calculation proceeds, and then specification of the subgroup H whose normalizer is to be calculated in the larger group G . Maple answers by giving a set of generators for N ( H ).

> normalizer(G, H);

permgroup(4,{[[1, 2, 3]], [[2, 3]]})


Notice that the normalizer of
H is actually a copy of the group S[3] in which we have seen both directly and via Maple that < ( 1 2 3 ) > is normal.