Quelques développements sur les groupes finis

G G G
C8G
C3
2
G a
G s 6=a2< a >
sas−1=a sas−1=a−1
sas−1=a G C4×C2sas−1=a−1G
D4
a2G G =< a, b > b
bab−1=a−1aba−1=b−1
G
G=<(1,2,5,4)(3,8,6,7),(1,8,5,7)(2,3,4,6) >S8
HG={1,−1, i, −i, j, −j, k, −k}
H?
G n2n3
G
n2n3
n2= 3 n3= 4
n2= 1 n3= 1 G C4×C3'C12 C2×C2×C3'
C2×C6
n2= 1 n3= 4 H G P 'C3
G G =HoP H H ⊂ZG(H)
ZG(H)H G ZG(H)
ZG(H)P H G
ZG(H) = H
ϕ:P−→ (H)
p−→ ϕ(p):(h→php−1)
(H)
(C4)'C2(C2×C2)'S3

H G
G G
ψ:G−→ SG/P 'S4G G 'A4
n2= 3 n3= 1 P=<t>'C3G
G Q =<
u, v >
ut =tu vtv−1=t−1
utu−1=t−1vtv−1=t−1uvtv−1u−1=ut−1u−1=t
u uv
s=ut =tu K =< u, t >=<s>
G G =Ko< v > vsv−1=vutv−1=ut−1=s−1
G D6
G
Q=< v > G vtv−1=t−1G
W=< t, v > t v
vtv−1=t−1
Gn(C)G
G N ≥2gN=e g ∈G
XN−1∈C[X]g g g ∈G
g∈G N
T={(g)/g ∈G}
{g1,· · · , gd} ⊂ Gn(C)G
τ:G−→ Td
g−→ ( (ggi))1≤i≤d
τ G
g, g0∈G τ(g) = τ(g0) (ggi) = (g0gi) 1 ≤i≤d
h∈G h =
d
P
i=1
cigi(gh) = (g0h)
h=g0−1(gg0−1) = n k ≥2h= (gg0−1)k−1g0−1
((gg0−1)k) = ((gg0−1)k−1)
((gg0−1)k) = n k ≥1
k≥0
((gg0−1−e)k) = (
k
X
i=1
Ci
k(gg0−1)i(−1)k−i=n
k
X
i=1
Ci
k(−1)k−i= 0
gg0−1−e gg0−1−e gg0−1∈G
gg0−1−e= 0 g=g0τ G

Gn(Z)
p6= 2 G
π:n(Z)−→ n(Z/Zp)
G N gN=e g ∈G
XN−1∈C[X]g g g ∈G
g∈G N
g∈G g =e+ph h ∈n(Z)
χgg
χg(X) = (Xe −g) = ((X−1)e−ph) = pnχh(1
p(X−1))
χg= (X−1)ng
g g =e
P, Q ∈Z[X]n λ P |λ|= 1
P(X) = pnQ(1
p(X−1)) P= (X−1)n
P=Q
λ
(X−λ)P(1) = Q
λ
(1 −λ) = pnQ(0) Q(0)
|P(1)|=pn|Q(0)| ≤ 2np≥3|Q(0)| ≤ (2
p)nQ(0) = P(1) = 0 P=
(X−1)P Q =XQ P (X) = pn−1Q(1
p(X−1))
K q
(n(K)) = (qn−1) · · · (qn−qn−1)
g∈n(K)C1(g)g Kn
qn−1C1(g)C2(g)g
KnC1(g) (qn−1) −(q−1) = qn−q)
C1(g)C2(g)C3(g)
KnKn(C1(g), C2(g)) qn−q2
G G n(Z)
(G)≤(3n−1) · · · (3n−3n−1)
p= 3
(n(Z/Z3)) = (3n−1) · · · (3n−3n−1)
G2(Z)
G3(Z)

≤59
G n < 60
G n ≥59
< n
n=prp r = 1
G r ≥2G
p
(G)
( (G)) ≡(G)p
(G)6={e}p
n=pq p < q
nq=kq + 1|p nq= 1 Q q G
Q'CqQ G G/Q 'Cppq
n=pqr p < q < r G
nr=λr + 1|pq ⇒nr=pq
nq=µq + 1|pr ⇒nq=r pr
np=νp + 1|qr ⇒np=q, r qr
(G)≥1 + pq(r−1) + p(q−1) + q(p−1) = 1 + pqr + (pq −(p+q))
| {z }
≥0
> pqr
G
n=p2q
p > q np=kp + 1|q np= 1 G
p2≤nq=kq + 1|p2nq= 1
n= 12 = 223n2
n3n2= 3 n3= 4 n3= 4
n= 45 = 325n5= 1
n=p3q p > q n = 40 = 235n5= 1
56 = 237n7= 1
n= 24
n= 24 n2= 2k+ 1|3n2= 1 3 n2= 1
n2= 3 P1P2G
(P1.P2) = (P1) (P2)
(P1∩P2)=64 (P1∩P2)≤24
8
3≥(P1∩P2)|8

(P1∩P2)=4
G=< P1, P2>
P1∩P2P1P2P1P2
G
n= 48 = 243n2= 2k+ 1|3n2= 1 3 n2= 1
n2= 3 P1P2G
(P1.P2) = (P1) (P2)
(P1∩P2)=28 (P1∩P2)≤243
16
3≥(P1∩P2)|16
(P1∩P2)=8
G=< P1, P2>
P1∩P2P1P2P1P2
G
n= 36 = 2232n3= 3k+ 1|4n3= 1 4 n3= 1
n3= 4 P1P2G
(P1.P2) = (P1) (P2)
(P1∩P2)=81 (P1∩P2)≤36
9
4≥(P1∩P2)|9
(P1∩P2)=3
G=< P1, P2>
P1∩P2P1P2P1P2
G
A E/Q
=e2π i
nK=Q[]K/Qϕ(n)
GK/Q−→ (Z/Zn)?
σ−→ kσ
σ() = kσ