Mathematics 566
Homework (due Halloween)
A. Hulpke
D&F;5.1: 1,4,7∗
D&F;5.2: 1,4,9∗
D&F;5.4: 7∗ ,15
D&F;5,5: 1
1) Determine (up to isomorphism) the abelian groups of order 2800 that can be generated by 3
elements.
2) Construct a non-abelian group of order 75.
The GAP command SemidirectProduct works as in the following example:
We want to construct the semidirect product of Z5 with Z4 (remember that Aut(Z5 ) ∼
= Z4 ). Both groups
have to be represented as permutation groups.
gap> n:=CyclicGroup(IsPermGroup,5);
Group([ (1,2,3,4,5) ])
gap> g:=CyclicGroup(IsPermGroup,4);
Group([ (1,2,3,4) ])
gap> actiongen1:=GroupHomomorphismByImages(n,n,[n.1],[n.1ˆ4]);
[ (1,2,3,4,5) ] -> [ (1,5,4,3,2) ]
gap> n.1; # the generator of n
(1,2,3,4,5)
gap> grpauts:=Group(actiongen1); # a group of automorphisms
<group with 1 generators>
gap> hominaut:=GroupHomomorphismByImages(g,grpauts,[g.1],[actiongen1]);;
gap> s:=SemidirectProduct(g,hominaut,n); # in some permutation representation
Group([ (2,5)(3,4)(6,7,8,9), (1,2,3,4,5) ])
3)∗ Construct in GAP the semidirect product of N = Z23 with G = GL2 (3) where G acts on N in the
natural way, interpreting N as a two-dimensional vector space over F3 :
a) Construct N as a permutation group
b) Construct G as a matrix group and use IsomorphismPermGroup to get an isomorphic permutation
group GP as its image.
c) For each generator of GP , construct (from the corresponding matrix) the automorphism of N as a
group homomorphism.
d) Construct the semidirect product as in the example above.
It is sufficient if you list your input, you do not need to copy lengthy permutations.
You may use GAP for any calculations. Problems marked with a ∗ are bonus problems.