¨Ubungsblatt 3, Abgabe am 20. Nov. 2015, um 09:00

Übungen zur Einführung in die Algebra, L. Hille
Wintersemester 2015/16
Übungsblatt 3, Abgabe am 20. Nov. 2015, um 09:00
Aufgabe 1. (3 Pt. + 1 Pt. Klarheit) Sei p eine Primzahl und G eine Gruppe mit ord G = pn .
Zeigen Sie mit Hilfe der Klassengleichung, dass p | ord Z(G).
Lösung. [1 Pt.] By the Klassengleichung,
ord (G) = ord (Z(G)) +
r
X
ord (K(gi )) = ord (Z(G)) +
i=1
r
X
(G : Zgi )
i=1
[1 Pt.] By definition, ord (G) = pn and, by the Lagrange’s Theorem, (G : Zgi ) = pki for some
ki > 0. In particular, p divides ord (G) andP
each (G : Zgi ).
r
[1 Pt.] In particular p divides ord (G) − i=1 (G : Zgi ) = ord (Z(G)).
Aufgabe 2. Sei p eine Primzahl, G eine Gruppe. Zeigen Sie:
(1) Wenn ord G = p2 , dann ist G abelsch und isomorph zu Z/p2 Z oder zu Z/pZ × Z/pZ.
(2) Wenn ord G = p3 , dann ist G abelsch oder es gilt ord Z(G) = p.
Lösung. 1) [1 Pt.] By the previous exercise, Z(G) has order p or p2 . If it has order p2 then
Z(G) = G and G is abelian.
[1 Pt.] We prove now that Z(G) cannot have order p. By contradiction, if it has order p,
then by Lagrange G/Z(G) has order p and therefore it is cyclic: in fact, taking any element
g ∈ G/Z(G), g 6= e, the order of the subgroup hgi must divide p but then it must be p and
therefore G/Z(G) = hgi is cyclic. However, by a previous exercise, if G/Z(G) is cyclic, then G is
abelian, and therefore Z(G) = G has order p2 , contradiction.
[1 Pt.] Therefore, Z(G) must have order p2 , and Z(G) is abelian. Given any element h ∈ G, the
order of h is either p or p2 . If there is an element g ∈ G of order p2 , then hgi = Z(G) and therefore
G is cyclic, equal to Z/p2 Z. If every element of G has order p, then take h1 ∈ G, and h2 ∈ G such
that h2 ∈
/ hh1 i. Define H1 = hh1 i, and H2 = hh2 i. Then H1 and H2 are commuting subgroups
of G of order p, and H1 ∩ H2 is a group of order 1 (it must divide p2 but it is smaller than p).
Therefore, H1 ∩ H2 = {e}, in particular the group H1 · H2 has order p2 : In fact, H1 H2 contains the
p2 elements of the form hi1 hj2 , 0 ≤ i, j ≤ p − 1, and all these elements must be different from each
= h2d−b and therefore this element belongs to H1 ∩ H2 .
other, ortherwise if ha1 hb2 = hc1 hd2 then ha−c
1
Therefore, it must be a = c and b = d. This also proves that, in this case, the map
φ : H1 × H2 → H1 H2 ,
φ(h1 , h2 ) = h1 h2
is an isomorphism.
Finally, since H1 H2 has order p2 , then G = H1 H2 ' H1 × H2 , but since H1 , H2 have order p,
they are both isomorphic to Z/pZ.
2) [1 Pt.] by the previous exercise, Z(G) has order p, p2 or p3 . If it is p3 , then G = Z(G) is
abelian. It cannot be p2 , because otherwise G/Z(G) is cyclic, and therefore G is abelian (same as
previous point), but then Z(G) = G must have order p3 . Therefore, either Z(G) has order p, or it
has order p3 in which case G is abelian.
1
Aufgabe 3. Sei G eine Gruppe, p eine Primzahl, und Syp die Menge die p-Sylow Untergruppen.
Zeigen Sie, dass ord Syp = (G : NG (S)), mit S eine Sylow Untergruppen und NG (S) = {g ∈ G |
gSg −1 ⊆ S}. (Hinweis: Benutzen Sie die Sylowsätze und die Formel ord (Gx) = ord (G)/ord (Gx )
für die Wirkung von G auf einer Menge X).
Lösung. Take the action µ : G × Syp → Syp , µ(g, H) = gHg −1 . This action is well-defined,
because if H is a Sylow p-subgroup of G, so is gHg −1 . By the first Sylow Theorem, the action
is transitive, in the sense that for any Sylow group S, the Bahn through S is the entire set Syp :
G · S = Syp . Moreover, the isotropy group at some S ∈ Syp is, by definition, the set NG (S):
GS = NG (S). From the general formula ord (G · p) = (G : ord (Gp )), valid for every action, it
follows in this case
ord Syp = ord (G · S) = (G : ord (GS )) = (G : NG (S))
Aufgabe 4. Sei G = S4 . Zeigen Sie:
(1) die Untergruppe
P = {e, (1234), (13)(24), (1432), (24), (14)(23), (13), (12)(34)}
ist eine 2-Sylow Untergruppe von S4 , und dass ist P kein Normalteiler.
(2) ord Sy2 = 3.
Lösung. 1)ord (S4 ) = 24 = 23 · 3, therefore every subgroup has order 8. P is a subgroup of
order 8, therefore it is a Sylow subgroup. It cannot be normal: for example, π = (24) ∈ P , but
taking σ = (12), we have
σπσ −1 = (12)(24)(12) = (14) ∈
/P
2) From the previous point, ord Sy2 = (G : NG (P )). From the definition, it follows that P ⊆
NG (P ) and therefore ord NG (P ) ≥ 8. Since NG (P ) is a subgroup of G = S4 , its order must divide
24. But since it contains P as a subgroup, its order must also be a multiple of 8. In particular,
ord (NG (P )) is either 8 or 24. However, if it was 24, it would be NG (P ) = G = S4 , which would
mean that, for every σ ∈ S4 , σP σ −1 = P , which is equivalent to saying that P is a normal
subgroup. By the previous point P is not a normal subgroup, therefore NG (P ) has order 8, and
therefore
ord Sy2 = (G : NG (P )) = ord (G)/ord (NG (P )) = 24/8 = 3.

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