**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
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);**

Could it be that the cyclic subgroup
*H*
generated by the 3-cycle ( 1 2 3 ) is
*always*
normal in
, no matter the value of
*n*
? To investigate, let's ask Maple whether the subgroup generated in
by that 3-cycle ( 1 2 3) is normal. Note that
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);**

Maple's answer kills the conjecture about
*H*
being normal in
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
. 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);**

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