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Groups With No Normal Subgroups of Order 9

The document summarizes the possible group structures of groups of order 36. There are two cases - groups with no normal subgroup of order 9, and groups with a normal subgroup of order 9. For groups with no normal subgroup of order 9, there are two possible structures: either the direct product of C3 and A4, or a semidirect product of C9 and a Klein group C22. For groups with a normal subgroup of order 9, there are four possible cases involving semidirect products of cyclic groups of orders 9 and 4 or 22.

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0% found this document useful (0 votes)
220 views3 pages

Groups With No Normal Subgroups of Order 9

The document summarizes the possible group structures of groups of order 36. There are two cases - groups with no normal subgroup of order 9, and groups with a normal subgroup of order 9. For groups with no normal subgroup of order 9, there are two possible structures: either the direct product of C3 and A4, or a semidirect product of C9 and a Klein group C22. For groups with a normal subgroup of order 9, there are four possible cases involving semidirect products of cyclic groups of orders 9 and 4 or 22.

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aye pyone
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© © All Rights Reserved
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Groups of order 36

Groups with no normal subgroups of order 9


Let G be a group of order 36 with no normal subgroup of order 9. If
P is any subgroup of order 9, the action of G on the set of left cosets G/P
defines a homomorphism

φ : G −→ Sym(G/P) ' S4 . (1)

The order of φ(G) divides both |Sym(G/P)| = 24 and |G| = 36, and is di-
visible by |G/P| = 4 since G acts transitively on G/P. This leaves us with
only two possibilities: |φ(G)| = 4, or 12. In the former case, Ker φ would
be a normal subgroup of order 9, hence contradicting the hypothesis. Thus,
|φ(G)| = 12, and Ker φ is a normal subgroup of order 3.
The alternating group A4 does not admit a nontrivial homomorphism
into a group of order 2. Since |Sym(G/P)/φ(G)| = 2, the composite homo-
morphism
Alt(G/P) ,→ Sym(G/P) 9 Sym(G/)/φ(G) (2)
is trivial showing that φ(G) coincides with Alt(G/P) ' A4 .
We have thus established that G fits into an extension
π
C3 v w G w w A4 . (3)

The adjoint action of G on Ker φ induces a homomorphism from A4 '


G/C3 to the group Aut C3 whose order equals 2. This being trivial, implies
that G acts trivially on C3 . In other words, C3 is contained in the center
Z(G). Since Z(G)/C3 ⊆Z(G/C3 ) ' Z(A4 ) = 1, we infer that Z(G) = C3 .
Group A4 has a unique subgroup W of order 4 and the latter is of the
form C22 . Since Ker π is central of order 3, for any w ∈ W, the cube w
e 3 of
an arbitrary lift w −1
e ∈ π (w) depends only on w, and defines a well defined
map
s : V −→ G (4)
which splits π. In particular,
3 s(w) is the only element v ∈ G such that
π(v) = w and v2 = w e 2 = 1. This immediately implies that map (4) is
a homomorphism embeding W into G, and that V ˜ s(W) is the unique
subgroup of order 4 in G.

1
In particular, V is normal and the abelian group C3 V ' C3 ×V is a normal
subgroup of G of order 12.
Case 1: G has at least three elements of order 3. In this case,
at least one of them, u, does not belong to center C3 . The subgroup H ˜
Vhui⊂G, which has exactly 12 elements, is generated by the set X ˜ V ∪ {u},
and its image π(X) generates group A4 that has the same number of elements
as H. It follows, that π restricted to H is an isomorphism of H with A4 . Since
|G| = |C3 H|, and C3 is central, we conclude that

G = C3 H ' C3 × A4 . (5)

Case 2: G has exactly two elements of order 3. In this case, any


lift to G of any element of order 3 in A4 has order 9. In particular, by
considering the cyclic subgroup of G generated by any element t of order
9, we obtain a representation of G as a nontrivial1 semidirect product of a
cyclic group of order 9 and the Klein group:

G = C9 V = C9 n V ' C9 n C22 . (6)

Since there are only two nontrivial homomorphisms

ρi : C9 −→ Aut C22 ' S3 , (i = 1, 2), (7)

and ρ2 (t) = ρ1 (t−1 ), there exists only one nonabelian semidirect product (6)
up to isomorphism. This is thus a unique nontrivial central extension of A4
by C3 .

In particular, we have proved that there are only two groups of order 36,
up to isomorphism, with no normal subgroup of order 9.
Note that in Case 1, there are exactly 3 elements of order 2 in G, 3 · 8 +
2 = 26 elements of order 3, 2 · 3 = 6 elements of order 6, and no elements of
order 9.
In Case 2, there are exactly 2 elements of order 3 in G and 4 · 6 = 24
elements of order 9. The number of elements of order 2 and 6 is the same
as in the other case.
The set of subgroups of order 9 is freely operated by the Klein group V.
1 Otherwise G would have a normal subgroup of order 9.

2
Groups with a normal subgroup of order 9
There are four cases:
Case 1: G ' C9 o C4 . There is only one nontrivial homomorphism

ρ : C4 −→ Aut C9 , m 7−→ ρm , (8)


m
where ρm sends a generator g ∈ C9 to g(−1) . Thus, in this case G is either
abelian and then cyclic, or nonabelian, and then isomorphic to the unique
nontrivial central extension
π
C2 v w G w w D9 (9)

of dihedral group D9 by C2 .
Case 2: G ' C9 o C22 . There is only one, up to an automorphism of C22 ,
nontrivial homomorphism

ρ : C22 −→ Aut C9 , (m, n) 7−→ ρ(m,n) , (10)

where ρ(1,0) sends a generator g ∈ C9 to g−1 while ρ(0,1) = idC9 . Thus, in


this case G is either abelian and then isomorphic to C18 ×C2 , or nonabelian,
and then isomorphic to D18 ' D9 × C2 .
3
Case 3: G ' C3 o C4 .
3
Case 4: G ' C3 o C22 . In this and the previous case G is the semidirect
product of a Sylow 2-subgroup Q with any two-dimensional representation
of Q over the three-element field F3 .

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