Generalized Quaternions

1. introduction
The quaternion group Q8 is one of the two non-abelian groups of size 8 (up to isomorphism). The other one, D4 , can be constructed as a semi-direct product:
D4 ∼
= Aff(Z/(4)) ∼
= Z/(4) o (Z/(4))× ∼
= Z/(4) o Z/(2),
where the elements of Z/(2) act on Z/(4) as the identity and negation. While Q8 is not a
semi-direct product, it can be constructed as the quotient group of a semi-direct product.
We will see how this is done in Section 2 and then jazz up the construction in Section 3 to
make an infinite family of similar groups with Q8 as the simplest member. In Section 4 we
will compare this family with the dihedral groups and see how it fits into a bigger picture.
2. The quaternion group from a semi-direct product
The group Q8 is built out of its subgroups hii and hji with the overlapping condition
i2 = j 2 = −1 and the conjugacy relation jij −1 = −i = i−1 . More generally, for odd a we
have j a ij −a = −i = i−1 , while for even a we have j a ij −a = i. We can combine these into
the single formula
j a ij −a = i(−1)
for all integers a. These relations suggest the following way of building the quaternion group
from scratch.
Theorem 2.1. Let H = Z/(4) o Z/(4), where
(a, b)(c, d) = (a + (−1)b c, b + d),
The element (2, 2) in H has order 2, lies in the center, and H/h(2, 2)i ∼
= Q8 .
Proof. Since −2 = 2 in Z/(4),
(a, b)(2, 2) = (a + (−1)b 2, b + 2) = (a + 2, b + 2)
(2, 2)(a, b) = (2 + (−1)2 a, 2 + b) = (2 + a, b + 2) = (a + 2, b + 2),
(2, 2) is in the center of H. Also (2, 2)(2, 2) = (2 + (−1)2 2, 2 + 2) = (0, 0), so (2, 2) has order
2 in H. Therefore the quotient group
Q := H/h(2, 2)i
makes sense and has size 16/2 = 8.
Since H is generated by (1, 0) and (0, 1):
(a, b) = (a, 0)(0, b) = (1, 0)a (0, 1)b .
In Q, set i to be the class of (1, 0) and j to be the class of (0, 1), so i and j generate Q:
(a, b) = ia jb .
To create an isomorphsm Q → Q8 , we will back up and create a homomorphism from
the semi-direct product H onto Q8 and check (2, 2) is in its kernel, so we get an induced
homomorphism from Q onto Q8 .
Define f : H → Q8 by f (a, b) = ia j b . This is well-defined since i4 = 1 and j 4 = 1. It is a
homomorphism since
f ((a, b)(c, d)) = f (a + (−1)b c, b + d) = ia+(−1) c j b+d
f (a, b)f (c, d) = ia j b ic j d = ia (j b ic j −b )j b+d = ia (j b ij −b )c j b+d = ia i(−1) c j b+d ,
which are the same. The image of f is a subgroup of Q8 containing i = f (1, 0) and
j = f (0, 1), so the image is Q8 : f is onto. Since f (2, 2) = i2 j 2 = (−1)(−1) = 1, the kernel
of f contains (2, 2), so f induces a surjective homomorphism Q → Q8 given by ia jb 7→ ia j b .
The groups Q and Q8 have the same size, so this surjection is an isomorphism.
3. Generalized Quaternions
While Q8 is made out of two cyclic groups of order 4, we can extend the construction by
letting one of the two cyclic groups be any cyclic 2-group.
Definition 3.1. For n ≥ 3, set
Q2n = (Z/(2n−1 ) o Z/(4))/h(2n−2 , 2)i,
where the semi-direct product has group law
(a, b)(c, d) = (a + (−1)b c, b + d).
The groups Q2n are called generalized quaternion groups.
Note Q2n is not the semi-direct product Z/(2n−1 ) o Z/(4), but rather the quotient of this
semi-direct product modulo the subgroup h(2n−2 , 2)i. Since 2n−2 mod 2n−1 and 2 mod 4
have order 2 in the additive groups Z/(2n−1 ) and Z/(4), a calculation shows (2n−2 , 2) is in
the center of the semi-direct product and has order 2, so h(2n−2 , 2)i is a normal subgroup of
the semi-direct product and the size of Q2n is (2n−1 · 4)/2 = 2n . The next theorem brings
the construction of Q2n down to earth.
Theorem 3.2. In Q2n , let x = (1, 0) and y = (0, 1). Then Q2n = hx, yi, where
(1) x has order 2n−1 , y has order 4,
(2) every element of Q2n can be written in the form xa or xa y for some a ∈ Z,
(3) x2
= y2 ,
(4) for any g ∈ Q2n such that g 6∈ hxi, gxg −1 = x−1 .
This theorem says, roughly, that Q2n is made by taking a cyclic group of order 2n−1
and a cyclic group of order 4 and “gluing” them at their elements of order 2 while being
Proof. Since Z/(2n−1 ) is generated by 1 and Z/(4) is generated by 1, Q2n is generated by
the cosets of (1, 0) and (0, 1), so x and y generate Q2n .
(1): The smallest power of (1, 0) in h(2n−2 , 2)i = {(2n−2 , 2), (0, 0)} is its 2n−1 -th power,
which is (0, 0), so x has order 2n−1 in Q2n . Similarly, the smallest power of (0, 1) in
h(2n−2 , 2)i is its fourth power, so y has order 4 in Q2n .
(2) and (3): A typical element of Z/(2n−1 ) o Z/(4) has the form (a, b) = (1, 0)a (0, 1)b ,
so every element of Q2n has the form xa yb . Since (2n−2 , 2) is trivial in Q2n and (2n−2 , 2) =
(1, 0)2 (0, 1)2 , x2
= y−2 = y2 in Q2n . Therefore in a product xa yb we can absorb any
even power of y into the power of x, which means we can take b = 0 or b = 1.
(4): Any g 6∈ hxi has the form g = xa y, so gxg −1 = xa yxy−1 x−a . Therefore it suffices
to focus on the case g = y. In Z/(2n−1 ) o Z/(4), (0, 1)(1, 0)(0, 1)−1 = (−1, 1)(0, −1) =
(−1, 0) = (1, 0)−1 , so yxy−1 = x−1 .
Since n ≥ 3, x has order greater than 2, so the condition yxy−1 = x−1 6= x, shows Q2n
is noncommutative. While we didn’t define Q2n when n = 2, the definition of Q4 makes
sense and is a cyclic group of order 4 generated by y (with x = y2 ).
The following theorem describes a special mapping property of Q2n : all groups with a
few of the basic features of Q2n are homomorphic images of it.
Theorem 3.3. For n ≥ 3, let G = hx, yi where x2
= 1, y 4 = 1, yxy −1 = x−1 , and
= y 2 . There is a unique homomorphism Q2n → G such that x 7→ x and y 7→ y, and
it is onto. If #G = 2n this homomorphism is an isomorphism.
The trivial group fits the conditions of the theorem (taking x = 1 and y = 1), so not all
such groups must be isomorphic to Q2n (only such groups of the right size are). Remember:
saying x2
= 1 and y 4 = 1 does not mean x has order 2n−1 and y has order 4, but only
that their orders divide 2n−1 and 4.
Proof. If there is a homomorphism Q2n → G such that x 7→ x and y 7→ y, then the
homomorphism is completely determined everywhere since x and y generate Q2n . So such
a homomorphism is unique. To actually construct such a homomorphism (prove existence,
that is), we adapt the idea in the proof of Theorem 2.1: rather than directly write down a
homomorphism Q2n → G, we back up and start with a map out of a semi-direct product
to G.
Let f : Z/(2n−1 ) o Z/(4) → G by f (a, b) = xa y b . This is well-defined since x2
and y = 1. To check f is a homomorphism, we will use the condition yxy = x , which
implies y b xy −b = x(−1) . First we have
f ((a, b)(c, d)) = f (a + (−1)b c, b + d) = xa+(−1) c y b+d ,
and next we have
f (a, b)f (c, d) = xa y b xc y d = xa (y b xc y −b )y b+d = xa (y b xy −b )c y b+d = xa x(−1) c y b+d .
Thus f is a homomorphism. It is surjective since we are told x and y generate G and x and
y are values of f . Since f (2n−2 , 2) = x2 y 2 = y 2 y 2 = y 4 = 1, (2n−2 , 2) is in the kernel of
f . Therefore f induces a surjective homomorphism Q2n → G given by xa yb 7→ xa y b , so G
is a homomorphic image of Q2n .
When #G = 2n , f is a surjective homomorphism between groups of the same size, so it
is an isomorphism.
Theorem 3.3 is true when n = 2 if we define Q4 = hyi to be a cyclic group of order 4
with x = y2 . Theorem 3.3 also gives us a recognition criterion for generalized quaternion
groups in terms of generators and relations.
Example 3.4. Let ζ = e2πi/2 , a root of unity of order 2n−1 . In GL2 (C), the matrix
ζ 0
0 1 ) has order 4. Since x2n−2 = y 2 = ( −1 0 )
x = ( 0 ζ ) has order 2n−1 and the matrix y = ( −1
0 −1
and yxy −1 = x−1 , the group generated by x and y is a homomorphic image of Q2n by
Theorem 3.3. Therefore hx, yi has size dividing 2n . This group contains hxi, of order 2n−1 ,
so 2n−1 |#hx, yi. We have y 6∈ hxi since x and y do not commute (because yxy −1 = x−1 6= x),
so #hx, yi = 2n . Therefore hx, yi ∼
= Q2n . The division ring of real quaternions a+bi+cj +dk
is isomorphic to the ring of complex matrices of the form ( wz −w
z ), where z = a + bi and
w = c + di. The matrices x and y have this form, so all the groups Q2n can be embedded
in the real quaternions.
Remark 3.5. The basic idea behind the construction of Q2n can be pushed further. Let A
be an abelian group of even order, written additively, and m be a positive integer divisible
by 4. There is an element of A with order 2. Pick one, say ε. Consider the semi-direct
product G = A o (Z/(m)), with group law as in (3.1). Since −ε = ε and m/2 is even, a
short calculation shows (ε, m/2) is in the center of G and has order 2. The quotient group
G/h(ε, m/2)i
generalizes the construction of Q2n (which is the special case A = Z/(2n−1 ), m = 4). This
group is noncommutative as long as A has an element of order greater than 2. When A
is cyclic of even order 2r (but not necessarily a 2-group, e.g., A = Z/(6) for r = 3) and
m = 4, this group is called a dicyclic group. It has order 4r with generators x and y such
that x2r = 1, yxy −1 = x−1 , and xr = y 2 .
4. Dihedral and Generalized Quaternion Groups
For all n ≥ 3, the dihedral group D2n−1 and quaternion group Q2n , both of order 2n ,
are similar in a number of ways. First, their generators and relations are analogous (but of
course not the same): D2n−1 = hr, si where
= 1,
s2 = 1,
srs−1 = r−1
and Q2n = hx, yi where
= 1,
y4 = 1,
yxy−1 = x−1 ,
= y2 .
The condition y4 = 1 can be dropped, since the first and last conditions imply it, but
we included it to make the similarity with dihedral groups clearer. In the degenerate case
n = 2, D2 ∼
= Z/(2) × Z/(2) and Q4 ∼
= Z/(4) are the two groups of size 4.
We will state without proof a catch-all theorem about the groups D2n−1 and then see
what the analogue is for Q2n .
Theorem 4.1. For n ≥ 3, D2n−1 has the following properties:
(1) the subgroup hri has index 2 and every element of D2n−1 outside of hri has order 2,
(2) the center of D2n−1 is {1, r2 } and D2n−1 /Z(D2n−1 ) ∼
= D2n−2 ,
(3) the commutator subgroup of D2n−1 is hr i, and D2n−1 /hr2 i ∼
= Z/(2) × Z/(2),
(4) there are 2n−2 +3 conjugacy classes, with representatives given in the following table.
Rep. 1 r r2 · · · r2 −1 r2
Size 1 2 2 · · ·
2n−2 2n−2
Table 1. Conjugacy class representatives in D2n−1
Theorem 4.2. For n ≥ 3, Q2n has the following properties:
the subgroup hxi has index 2 and every element of Q2n outside of hxi has order 4,
the center of Q2n is {1, x2 } = {1, y2 } and Q2n /Z(Q2n ) ∼
= D2n−2 ,
the commutator subgroup of Q2n is hx2 i, and Q2n /hx2 i ∼
× Z/(2),
there are 2
+3 conjugacy classes, with representatives given in the following table.
Rep. 1 x x2 · · · x2 −1 x2
Size 1 2 2 · · ·
2n−2 2n−2
Table 2. Conjugacy class representatives in Q2n
Proof. (1): Since x has order 2n−1 , [Q2n : hxi] = 2. The elements of Q2n that are not
powers of x have the form xa y, and
(xa y)2 = xa (yxa y−1 )y2 = xa (yxy−1 )a y2 = xa x−a y2 = y2 ,
so xa y has order 4.
(2): Since x2
= y 2 , x2
commutes with both x and y, and hence with all of Q2n ,
so x2
= y2 is in the center. If xa is in the center, then yxa y−1 = xa . The left side is
(yxy−1 )a = x−a , so x−a = xa . Therefore x2a = 1, so 2n−1 |2a, so 2n−2 |a, which means xa
is a power of x2 .
No element of Q2n that is not a power of x is in the center, since x(xa y)x−1 = xa+1 xy =
xa+2 y 6= xa y. (Here we need n ≥ 3 to be sure that x2 6= 1.)
The quotient group Q2n /Z(Q2n ) has generators x and y such that x2
= 1 (since
= y is in the center), y = 1, and yxy = x . Therefore this quotient group is a
homomorphic image of D2n−2 . Since the size of Q2n /Z(Q2n ) is 2n−1 = #D2n−2 , Q2n /Z(Q2n )
is isomorphic to D2n−2 : the cosets x and y in Q2n /Z(Q2n ) play the roles of r and s in the
dihedral group.
(3): Since xyx−1 y−1 = x2 , the commutator subgroup of Q2n contains hx2 i. (In fact,
x yx−a y−1 = x2a , so all elements of hx2 i are commutators.) The subgroup hx2 i has size
2n−2 and thus index 4. It is a normal subgroup of Q2n since yx2 y−1 = x−2 ∈ hx2 i. The
group Q2n /hx2 i has size 4, hence is abelian, so every commutator in Q2n is in hx2 i. Therefore
hx2 i is the commutator subgroup of Q2n . That Q2n /hx2 i ∼
= Z/(2) × Z/(2) is left to the
(4): For each g ∈ Q2n we will compute xa gx−a and (xa y)g(xa y)−1 = xa ygy−1 x−a as a
First suppose g is a power of x, say g = xk . Then
xa xk x−a = xk ,
(xa y)xk (xa y)−1 = x−k ,
so the conjugacy class of xk is {xk , x−k }, which has size 2 as long as x2k 6= 1. The two
exceptions here are g = 1 and g = x2 , which are in the center.
If g = y then
xa yx−a = x2a y, (xa y)y(xa y)−1 = x2a y,
so the conjugacy class of y is all x2a y as a varies.
Finally, if g = xy then
xa xyx−a = x2a+1 y,
(xa y)xy(xa y)−1 = x2a−1 y,
so the conjugacy class of xy is all x2a+1 y as a varies.
The first parts of Theorems 4.1 and 4.2 are a noticeable contrast between D2n−1 and Q2n :
at least half the elements of the dihedral group have order 2 and at least half the elements
of Q2n have order 4. The only elements of D2n−1 with order 4 are r2
and its inverse.
What are the elements of Q2n with order 2?
Corollary 4.3. The only element of Q2n with order 2 is x2
Proof. Since x has order 2n−1 , its only power with order 2 is x2
that is not a power of x has order 4 by Theorem 4.2(1).
. Any element of Q2n
Remark 4.4. While Theorem 4.2 lists some properties common to the groups Q2n for
all n ≥ 3, Q8 has a feature not shared by its higher analogues. In Q8 every subgroup is
normal, but for n ≥ 4 the group Q2n has the non-normal subgroup hyi = {1, y, y2 , y3 } =
{1, y, x2 , x2 y}. This is not normal because xyx−1 = x2 y, which does not belong to
hyi because 2 < 2n−2 .
Corollary 4.5. Any normal subgroup N C Q2n which is nonabelian has index 1 or 2.
Proof. Since N is nonabelian, N 6⊂ hxi. Pick g ∈ N with g 6∈ hxi. Then g has order 4
(Theorem 4.2(1)), so g 2 has order 2. That means g 2 = x2
(Corollary 4.3). Every element
of Q2 outside of hxi conjugates x into x (Theorem 3.2(4)), so gxg −1 = x−1 . Since N is
a normal subgroup of Q2n , N contains
g(xg −1 x−1 ) = (gxg −1 )x−1 = x−2 ,
so N ⊃ hx2 , gi. The subgroup hx2 i has index 4 in Q2n and N is strictly larger, so its index
is 1 or 2.
Corollary 4.6. For n ≥ 4, Aut(Q2n ) is a 2-group.
This is not true for n = 3: Aut(Q8 ) ∼
= S4 .
Proof. If Aut(Q2n ) is not a 2-group, it has an element f of odd order > 1. Set G = hxi.
All elements Q2n outside of G have order 4 (Theorem 4.2(1)), while G is cyclic of order
2n−1 > 4 (since n ≥ 4), so f must send any generator of G to another generator of G.
Therefore f (G) = G, so f restricts to an automorphism of G. Since G ∼
= Z/(2n−1 ),
Aut(G) ∼
)) , which is a 2-group (ϕ(2
), so f must be the identity on
= (Z/(2
G because f has odd order.
Since f (G) = G, also f (Q2n − G) = Q2n − G. Although Q2n − G is not a group, it is a
set and f is a permutation of this set. The size of Q2n − G is 2n − 2n−1 = 2n−1 , a power of
2, and f has odd order as a permutation on this set, so it must have a fixed point: f (q) = q
for some q ∈ Q2n − G. Then f is the identity on the subgroup hG, qi = hx, qi, which is Q2n
since G has index 2 in Q2n .
Here is an interesting role for the generalized quaternion groups, alongside cyclic p-groups.
Theorem 4.7. For a finite p-group, the following conditions are equivalent:
(1) there is a unique subgroup of order p,
(2) all abelian subgroups are cyclic,
(3) the group is cyclic or generalized quaternion.
Proof. See the appendix.
Corollary 4.8. Every subgroup of Q2n is cyclic or generalized quaternion.
Proof. There is a unique subgroup of order 2 in Q2n , so every nontrivial subgroup has a
unique subgroup of order 2. Now use Theorem 4.7.
Corollary 4.9. When p is an odd prime, a finite p-group is cyclic if and only if it has one
subgroup of order p. A finite 2-group is cyclic if and only if it has one subgroup of order 2
and one subgroup of order 4.
Proof. A generalized quaternion group Q2n has at least 2 subgroups of order 4, such as the
subgroup of order 4 in hxi and the subgroup hyi.
Corollary 4.10. If D is a division ring, any Sylow subgroup of a finite subgroup of D× is
cyclic or generalized quaternion.
Proof. Any finite abelian subgroup of D× is cyclic, so we can apply the second part of
Theorem 4.7 to its Sylow subgroups.
Remark 4.11. The classification of all finite groups in division rings was worked out by
Amitsur [2].
Theorem 4.7 is also applicable to the Sylow subgroups of finite groups with periodic
cohomology. (Groups with periodic cohomology arise in studying group actions on spheres.)
The finite groups with periodic cohomology were determined by Zassenhaus [6] for solvable
groups and by Suzuki [5] for non-solvable groups.
While Theorem 3.3 provides a criterion to recognize a generalized quaternion group in
terms of generators and relations, Theorem 4.7 provides a more abstract criterion: a noncyclic 2-group with a unique element of order 2 is generalized quaternion. Here is a nice
use of this, relying partly on Galois theory for finite fields.
Corollary 4.12. Let F be a finite field not of characteristic 2. The 2-Sylow subgroups of
SL2 (F ) are generalized quaternion groups.
Proof. (Taken from [3, p. 43].) The only element of order 2 in SL2 (F ) is ( −1
0 −1 ), so a 2Sylow subgroup of SL2 (F ) has a unique element of order 2. Therefore the 2-Sylow subgroup
is either a cyclic group or a generalized quaternion group. We need to eliminate the cyclic
option. It would be wrong to do this just by writing down two noncommuting elements
of 2-power order in SL2 (F ), because that by itself doesn’t imply the 2-Sylow subgroups
are noncommutative (and hence not cyclic): elements of 2-power order need not generate
a subgroup of 2-power order. For example, any dihedral group Dn = hr, si (n ≥ 3) is
generated by the two reflections s and rs, which both have order 2.
Let q = #F , so q is an odd prime power and # SL2 (F ) = q(q 2 − 1). We are going to
show every A ∈ SL2 (F ) with 2-power order has order dividing either q + 1 or q − 1. These
numbers are both even, so the highest power of 2 in # SL2 (F ) is not a factor of q + 1 or q − 1
and therefore the order of a 2-Sylow subgroup is not the order of any element of SL2 (F ).
Thus a 2-Sylow subgroup can’t be cyclic.
Since the characteristic polynomial of A has degree 2, its eigenvalues λ and µ are in F
or a quadratic extension of F , and λµ = 1 since det A = 1. If λ = µ either both eigenvalues
are 1 or both are −1, which would imply A is conjugate to either ( 10 1b ) or ( −1
0 −1 ), and
these have 2-power order only when b = 0, so A = ±I2 . The order of A is 1 or 2, which
both divide q + 1 and q − 1.
From now on we can assume λ 6= µ. Then A has distinct eigenvalues, so A is diagonalizable over a field containing its eigenvalues. We will show Aq−1 = I2 or Aq+1 = I2 . If the
eigenvalues of A are in F then A is conjugate over F to ( λ0 µ0 ), hence Aq−1 = I2 . If the
eigenvalues are not in F then the characteristic polynomial of A is irreducible over F , so λ
and µ are Galois conjugate over F . Therefore µ = λq by Galois theory for finite fields, so
1 = λµ = λq+1 and µq+1 = 1/λq+1 = 1. Since A is conjugate (over a quadratic extension of
F ) to ( λ0 µ0 ), Aq+1 = I2 .
A 2-Sylow subgroup of SL2 (F ) can be written down explicitly when q ≡ 1 mod 4. Let 2k
be the highest power of 2 in q − 1, so the highest power of 2 in q(q 2 − 1) = q(q − 1)(q + 1)
is 2k+1 . The group F × is cyclic of order q − 1, so it contains an element a with order
2k . Let x = ( a0 1/a
) and y = ( 01 −1
0 ). Both are in SL2 (F ), x has order 2 , y has order 4,
= −I2 = y 2 , and y 6∈ hxi, so hx, yi ∼
= Q2k+1 by the same argument as in Example 3.4.
In particular, hx, yi has order 2k+1 , so it is a 2-Sylow subgroup of SL2 (F ).
As an example of this, when q = 5 we can use a = 2: the 2-Sylow subgroup of SL2 (F5 ) is
h( 0 03 ), ( 01 −1
0 )i and is isomorphic to Q8 . For k ≥ 1, the highest power of 2 dividing 3 − 1
is k + 2, so the 2-Sylow subgroup of SL2 (F32k ) is isomorphic to Q2k+2 .
Alas, when q ≡ 3 mod 4 the group F × has no elements of 2-power order besides ±1,
since the highest power of 2 in q − 1 is 2. So the explicit construction above of a 2-Sylow
subgroup of SL2 (F ) no longer works. For a generator and relations method of showing the
2-Sylow subgroup of SL2 (F ) is generalized quaternion when q ≡ 3 mod 4, see [1, p. 147].
What if F has characteristic 2? Letting q = #F , which is a power of 2, the 2-Sylow
subgroups of SL2 (F ) have order q and {( 10 x1 ) : x ∈ F } is a subgroup of order q which is
isomorphic to the additive group of F . So a 2-Sylow in SL2 (F ) is a direct sum of cyclic
groups of order 2, since that is the group structure of F (additively).
What can be said about the structure of the p-Sylow subgroups of SL2 (F ) at odd primes
p? If p is the characteristic of F , then the p-Sylow subgroup is isomorphic to F by the same
argument used in the previous paragraph. If p is an odd prime dividing q 2 − 1 then the
p-Sylow subgroups of SL2 (F ) are cyclic, but we omit the proof.
Appendix A. Proof of Theorem 4.7
We will prove Theorem 4.7 following the argument in [4, Theorem 9.7.3].
We want to show a nontrivial finite p-group is cyclic or generalized quaternion if it has a
unique subgroup of order p or if all of its abelian subgroups are cyclic. (A nontrivial cyclic
p-group and a generalized quaternion group have both of these properties.)
We will consider separately abelian and non-abelian p-groups.
If G is a nontrivial abelian p-group, then by the structure theorem for finite abelian
groups we can write G as a direct product of cyclic p-groups: G ∼
= Z/(pr1 ) × · · · × Z/(prd ).
If d > 1 then G has more than one subgroup of order p and it has a non-cyclic subgroup
(such as G itself). Hence a finite abelian p-group is cyclic if it has a unique subgroup of
order p or if all of its (abelian) subgroups are cyclic.
From now on let G be non-abelian. If G has a unique subgroup of order p or if all of its
abelian subgroups are cyclic, we want to show G is a generalized quaternion group. If G has
a unique subgroup of order p then all of its nontrivial subgroups share this property, so all
of its abelian subgroups are cyclic by the previous paragraph. Therefore it suffices to focus
on the hypothesis of all abelian subgroups being cyclic, and show (when G is nonabelian)
this forces G to be generalized quaternion.
Since G is non-abelian, its center Z is a nontrivial proper subgroup of G, and Z has to
be cyclic since it’s abelian. For any g ∈ G, the subgroup hg, Zi is abelian, hence cyclic. The
subgroups of a cyclic p-group are totally ordered, so either hgi ⊂ Z or Z ⊂ hgi. Therefore
g 6∈ Z =⇒ Z $ hgi.
In particular, #Z is less than the order of g. So all elements of G − Z must have order
at least p2 . If p is odd we will construct an element of G − Z with order p, which is a
contradiction, so p = 2. If p = 2 we will construct an element of G − Z with order 4, so
#Z = 2.
Since G/Z is nontrivial, it contains an a with order p: a 6∈ Z and ap ∈ Z. Thus
hap i ⊂ Z $ hai by (A.1). Since hap i has index p in hai, and Z 6= hai, we must have
Z = hap i.
Subgroups of a cyclic p-group are totally ordered, so all proper subgroups of hai are in Z.
Since a 6∈ Z, some b ∈ G does not commute with a. Therefore hai ∩ hbi is a proper
subgroup of hai, so hai ∩ hbi ⊂ Z. At the same time, Z is a subgroup of hai and hbi by
(A.1), so
Z = hai ∩ hbi.
Since b 6∈ hai, hai ∩ hbi is a proper subgroup of hbi, so hai ∩ hbi = hbp i for some r ≥ 1.
Since hai ∩ hbi = Z = hap i, bp and ap generate the same group, so bp = apk for some k not
divisible by p. Since hak i = hai and hapk i = hap i, we can rename ak as a to have bp = ap
while still having
hai ∩ hbi = Z = hap i = hbp i.
Since Z = hap i ⊂ hai ∩ hbp i ⊂ hai and a ∈
/ hbp i (a and b do not commute), the second
inclusion is strict, so hai ∩ hbp i = Z. Now we rename bp
as b, so bp = ap and
hai ∩ hbi = Z = hap i.
Let c = b−1 , so a and c do not commute (recall a and b do not commute). Up to this
point, all we have used about a is that a has order p in G/Z. (In the course of the proof
we replaced a with a power ak such that (p, k) = 1, but this doesn’t change the condition
that a has order p.) Since G/Z is a nontrivial p-group, its center is nontrivial, so we could
have chosen a from the beginning such that a is an element of order p in the center of G/Z.
Make that choice. Then in G/Z, a and c commute, so
ca = acz
for some z ∈ Z with z 6= 1. Rewriting (A.2) as a−1 ca = cz and raising to the p-th power,
a−1 cp a = cp z p . Since cp = b−p = a−p , we obtain 1 = z p . From (A.2) and induction,
(ac)n = an cn z ( 2 )
for all positive integers n. Setting n = p in (A.3),
(ac)p = ap cp z p(p−1)/2 = ap a−p z p(p−1)/2 = z p(p−1)/2 .
If p 6= 2 then p is a factor of p(p − 1)/2, so z p(p−1)/2 = 1 because z p = 1. Thus (ac)p = 1.
Since c 6∈ hai, ac 6= 1, so ac has order p. But ac 6∈ hai ⊃ Z, so ac is an element of order p
in G − Z, which we noted earlier is impossible. Hence p = 2, so G is a 2-group and z 2 = 1.
Returning to (A.3) and setting n = 4,
(ac)4 = a4 c4 z 6 = a4 b−4 (z 2 )3 = 1,
so ac has order dividing 4. Since (ac)2 = a2 c2 z = a2 b−2 z = z 6= 1, ac has order 4. Since
ac 6∈ hai ⊃ Z, #Z < 4 by (A.1), so #Z = 2.
There is a normal subgroup N C G with order 4. It must be abelian, so it is cyclic.
Consider the conjugation action of G on N , which is a group homomorphism G → Aut(N ) ∼
{±1}. The center of G has order 2, while N has order 4, so not every element of G commutes
with every element of N , which means the conjugation action G → Aut(N ) is onto. Let
K be the kernel, so K has index 2 in G and thus is a normal subgroup of G. All abelian
subgroups of K are cyclic because it is so in G. Since #K < #G, by induction K is either
cyclic or generalized quaternion. Since N is abelian, N ⊂ K (look at the definition of K),
so N ⊂ Z(K). Then the center of K has size at least 4, which means K is not generalized
quaternion, so K is cyclic.
In the cyclic 2-group K there are two elements of order 4, which are inverses of each
other. If these are the only elements of G with order 4 then any element not of order 1 or
2 has these as powers of it, so commutes with them. The elements of order 1 or 2 commute
with everything since they are in the center of G, so the elements of order 4 in K commute
with everything. That means #Z(G) ≥ 4, a contradiction. Thus there has to be some
y ∈ G − K with order 4. Since y 6∈ K, y acts by conjugation nontrivially on N .
Set #G = 2n and K = hxi, so x has order 2n−1 , N = hx2 i, and Z = hx2 i. Since
the conjugation action of y on N is nontrivial,
y −1 = x−2
Since K C G, yxy −1 = xi for some i. We have G = hx, yi since K has index 2 and y 6∈ K,
so x and y don’t commute (G is nonabelian). Therefore yxy −1 6= x, so i 6≡ 1 mod 2n−1 .
We have y 2 ∈ K, since [G : K] = 2, and x2
is the only element of order 2 in K, so
y 2 = x2
Therefore y 2 xy −2 = y(yxy −1 )y −1 = yxi y −1 = (yxy −1 )i = xi , so i2 ≡ 1 mod 2n−1 . When
n = 3 we have i2 ≡ 1 mod 4 and i 6≡ 1 mod 4, so i ≡ −1 mod 4. Now let n ≥ 4. From
i2 ≡ 1 mod 2n−1 , we get i ≡ ±1 or 2n−2 ± 1 mod 2n−1 . We want to show i ≡ −1 mod 2n−1 ,
since then G ∼
= Q2n by Theorem 3.3. We know already that i 6≡ 1 mod 2n−1 , so it remains
to eliminate the choices i ≡ 2n−2 ± 1 mod 2n−1 .
Assume i ≡ 2n−2 ± 1 mod 2n−1 . Then xi = x2 ±1 = y 2 x±1 , so yxy −1 = y 2 x±1 .
xy −1 = yx±1 .
If xy −1 = yx−1 then xy −1 = (xy −1 )−1 , so (xy −1 )2 = 1. Elements of order 1 and 2 in G are
in Z ⊂ hxi, so xy −1 is a power of x. Thus y is a power of x, but x and y don’t commute.
We have a contradiction.
If xy −1 = yx then xy −1 x−1 = y. Conjugating by x again, x2 y −1 x−2 = xyx−1 = y −1 , so
x2 and y commute. Then x2 is in the center of G. But the center has order 2 and x2 has
order 2n−2 > 2, so we have a contradiction. Alternatively, we get a contradiction since the
subgroup hx2 , yi is abelian and not cyclic since hyi and hx2 i are two subgroups of it with
order 4.
A. Adem and R. J. Milgram, “Cohomology of Groups,” 2nd ed., Springer-Verlag, Berlin, 2004.
S. A. Amitsur, Finite subgroups of division rings, Trans. Amer. Math. Soc. 80 (1955), 361–386.
D. Gorenstein, “Finite Groups,” Harper & Row, New York, 1968.
W. R. Scott, “Group Theory,” Dover, New York, 1987.
M. Suzuki, On finite groups with cyclic Sylow subgroups for all odd primes, Amer. J. Math 77 (1955),
[6] H. Zassenhaus, Über endliche Fastkörper, Abh. Math. Sem. Hamburg Univ. 11 (1935), 187–220.
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