#Aufgabe 1
ChineseRem([54,80,225],[1,13,28]);
7453 mod 54;
7453 mod 80;
7453 mod 225;
Lcm(54,80,225);


#Aufgabe 2
G:=Group((1,2,3,4,5),(2,5)(3,4));
Order(G);
Elements(G);

l:=LatticeSubgroups(G);
c:=ConjugacyClassesSubgroups(l);
for sg in c do
Print(Order(sg[1])," ",Elements(sg),"\n");
od;

# Die Normalteiler sind
Elements(c[1]);
Elements(c[3]);
Elements(c[4]);

NormalSubGroups(G);

# Die 2-Sylowuntergruppen sind
Elements(c[2]);
# die 5-Sylowuntergruppe
Elements(c[3]);

# Die Konjugationsklassen von G
cc:=ConjugacyClasses(G);
for cl in cc do
Print(Size(cl)," ",Elements(cl),"\n");
od;

# Die Klassengleichung ist also
1+5+2+2;


# Aufgabe 3
G:=Group((1,2,3,4),(1,3));
Order(G);
Elements(G);
autG:=AutomorphismGroup(G);
Elements(autG);
innG:=InnerAutomorphismsAutomorphismGroup(autG);
Elements(innG);
Center(G);
IsomorphismGroups(G,autG);

gens:=[(1,2,3,4,5),(5,6)];
G:=Group(gens);
Size(G);
phi:=GroupHomomorphismByImages(G,G,gens,[(1,2,3,4,5),(1,2)(3,5)(4,6)]);
Kernel(phi);


# Aufgabe 4
G:=Group((2,3,4,5,6)(7,8,9,10,11),(1,2,8,9,4)(5,6,7,12,10),(3,6)(4,5)(7,8)(9,11));
Size(G);
cc:=ConjugacyClasses(G);
for cl in cc do
Print(Size(cl)," ","\n");
od;
1+1+15+15+12+12+12+12+20+20;


# Aufgabe 5
F := FreeGroup( "a", "b" );
G := F / [ F.1^4, F.2^6, (F.1*F.2)^2, (F.1^3*F.2^2)^2 ];
W:=Group((2,3,5,4),(1,5,3,6,2,4));
IsomorphismGroups(G,W);

# Aufgabe 6
G:=Group([(1,2,3,4)]);
H:=Group([(1,2)]);
AutG:=AutomorphismGroup(G);
elAutG:=Elements(AutG);
img:=elAutG[2];
phi:=GroupHomomorphismByImages(H,AutG,[(1,2)],[img]);
P:=SemidirectProduct( H,phi,G );
Elements(P);
D4:=Group((1,2,3,4),(1,2)(3,4));
IsomorphismGroups(P,D4);

gens:=[(1,2,3),(2,3,4)];
G:=Group(gens);
Size(G);
H:=Group([(1,2)]);
AutG:=AutomorphismGroup(G);
img1:=(1,2)*(1,2,3)*(1,2);
img2:=(1,2)*(2,3,4)*(1,2);
img:=GroupHomomorphismByImages(G,G,gens,[img1,img2]);
phi:=GroupHomomorphismByImages(H,AutG,[(1,2)],[img]);
P:=SemidirectProduct( H,phi,G );
Elements(P);
S4:=SymmetricGroup(4);
IsomorphismGroups(P,S4);