Abstract Algebra with Maple
Chapter 4: Finite Groups and Subgroups

4: Finite Groups and Subgroups

Orders of Elements

After loading the number theory package (we did it in the previous section), orders of elements of groups can be evaluated as follows:

> order(7,15);

> order(31,42);

The cyclic subgroup of generated by a , can be found using the following procedure:

> cycleU:=(c,n)->[seq(c^(i-1) mod n, i=1..order(c,n))]:

> cycleU(7,15);

> cycleU(31,42);

In general, the cyclic subgroup generated by an element c of an (extended) group g can be found using the following procedure:

> Cycle:=proc(c,g) local v,a;
v:=[g[3]]; a:=c;
while not a=g[3] do v:=[op(v),a]; a:=g[2](a,c) od; v end:

For example, the cyclic subgroup generated by 10 in :

> Cycle(10,Z(25));

Another example,

> map(matrix,Cycle([[1,2],[3,4]],GL2(5)));

The orders of elements can be found as the orders of the cyclic subgroups generated by them:

> Ord:=proc(c,g) local a,n;
a:=c; n:=1;
while not a=g[3] do n:=n+1; a:=g[2](a,c) od; n end:

> Ord([[1,2],[3,4]],SL2(5));

> Ord([[2,3],[4,5]],GL2(11));

> Ord(715,Z(1001));

Center and Centralizers

The centralizer of an element a of an extended group G can be determined as follows:

> CentralizerE:=proc(a,G) local i,v;
v:=[]; for i to nops(G[1]) do if G[2](G[1][i],a)=G[2](a,G[1][i]) then v:=[op(v),G[1][i]] fi od; v end:

For example,

> CentralizerE(8,Cyclic(8));

> CentralizerE(8,Dihedral(4));

The centralizer of a subset S of an extended group G can be determined as follows:

> CentralizerS:=proc(S,G) local i,j,v;
v:=[]; for i to nops(G[1]) do j:=1; while not j=nops(S)+1 and
G[2](G[1][i],S[j])=G[2](S[j],G[1][i]) do j:=j+1 od; if j=nops(S)+1 then v:=[op(v),G[1][i]] fi od; v end:

> CentralizerS([7,8],Dihedral(4));

The center is the centralizer of the group:

> Center:=G->CentralizerS(G[1],G):

> Center(Dihedral(7));

> Center(Dihedral(10));

> map(matrix,Center(GL2(3)));

Note.

Why do we need two commands for a centralizer - CentralizerE and CentralizerS ? Why can't we use one command, Centralizer, for both cases? The problem is that some elements of a group can be equal to some sets of other elements. For example, a group can contain elements 1, 2, and {1,2}.

What would the Centralizer({1,2}) be in that case? The centralizer of the element {1,2}, CentralizerE ({1,2}), or the centralizer of the subset {1,2}, CentralizerS ({1,2})? Since we can't distinguish between these two cases by checking the type of an argument, we need two different commands for that.

A Subgroup Test

According to Theorem 3.3 on p. 62 of Dr. Gallian's text, to test whether a finite subset H is a subgroup of G , it is enough to check if H is a subset of G and it is closed under the group operation of G . So we can use isCP for that:

> isSubgroup:=(h,G)->verify({op(h)},{op(G[1])},`subset`) and isCP(h,G[2]):

For example,

> isSubgroup(Cycle(8,Z(33)),Z(33));

> isSubgroup(Cycle([[1,2],[2,0]],GL2(5)),SL2(5));

Certainly, cyclic subgroups are subgroups :-) as well as the center and centralizers:

> isSubgroup(CentralizerE(11,Dihedral(6)),Dihedral(6));

> isSubgroup(CentralizerS({[[1,2],[3,4]],[[1,3],[2,4]]},GL2(5)),GL2(5));

> isSubgroup(Center(SL2(4)),SL2(4));

Finally, an opposite example:

> isSubgroup(Z(5)[1],Z(10));

>

Exercises

• 1 . Find the order of 7 in U (100) and U (11).
• 2 . Find the order of 6 in , , , , , and .
• 3 . Find the cyclic subgroup of U (145) generated by 19.
• 4 . Find the cyclic subgroup of GL (2, ) generated by .
• 5 . Find the order of cyclic subgroups of SL (2, ) and SL (2, ) generated by .
• 6 . Find the centralizer of 3 in the dihedral group .
• 7 . Find the centralizer of { , } in GL (2, ).
• 8 . Find the center of SL (2, ).
• 9 . Test if the set is a subgroup of U (45).