Finite Groups
Finite Groups
As a relation between number theory and
group theory, usually we get the Euler theorem
and Fermat theorem from Lagrange theorem
on finite groups.
Now we explore some more connections between numbers and finite groups.
First we recall some known examples and results of groups.
For any natural number n ≥ 2,
(ℤn = {0, 1, 2, . . . , n-1}, ⊕n) is a cyclic group of order n.
For any natural number n ≥ 3, the symmetric group of degree n,
Sn is a non-abelian group of order n!.
The group of symmetries of squares
D4 = {R0 = e, R90, R180, R270, H, V, D, D′}
is a group of order 8.
D4 has 4 rotational symmetries of rotation
by multiples of π/2 and 4 reflection symmetries,
reflection about the lines joining the midpoints
of opposite sides and the lines joining the opposite vertices.
D4 = {a0 = e, a, a2, a3, b, ab, a2b, a3b},
where a4 = e = b2 and ba = a3b = a-1b is a
nonabelian group of order 8.
Here a is the rotation by π/2 and
b is the reflection about the line through
the midpoints of any two opposite sides.
{a0 = e, a, a2, a3} is a cyclic subgroup
of D4, of order 4.
In general, symmetries of regular polygon
with n sides is a nonabelian group of
order 2n and it is denoted by Dn.
Dn has n rotational symmetries of rotation
by multiples of 2Ï€/n and n reflection symmetries.
If n is even, reflection about the lines joining
the midpoints of opposite sides and reflection
about the lines joining the opposite vertices
are the reflection symmetries.
If n is odd, reflection about the lines
joining each vertex with the midpoint
of the opposite side are the reflection symmetries.
Dn = {a0 = e, a, a2, . . . , an-1, b, ab, a2b, . . . , an-1b},
where an = e = b2 and
ba = an-1b = a-1b is a nonabelian group of order 2n.
Here a is the rotation by 2Ï€/n.
If n is odd b is the reflection about the line
through any one of the vertices and the midpoint
of the opposite side and if n is even
b is the reflection about the line through
the mid points of any two opposite sides.
Dn is called the dihedral group of order 2n.
{a0 = e, a, a2, . . . , an-1} is a cyclic subgroup
of Dn, of order n.
Dn can be considered as a subgroup of Sn.
D3 = S3 but Dn ≠ Sn, for all n ≥ 4.
(compare their orders)
Sn and Dn!/2 are non isomorphic,
nonabelian groups of order n!, for all n ≥ 4.
Dn!/2 has an element of order n!/2,
but Sn has no elements of order n!/2.
Cyclic subgroup of a group G generated by
an element a ∈ G,
<a> = {an / n ∈ ℤ }.
Any cyclic group is abelian.
Any group of prime order is cyclic.
Order of a cyclic group is the same as the
order of its generator.
Any infinite cyclic group is isomorphic to (ℤ, +).
Any cyclic group of order n is isomorphic to (ℤn, ⊕n).
Learning from the above results :
For any prime p, (ℤp, ⊕p) is the only group,
up to isomorphism, of order p and it is abelian.
(see how this follows from the above results)
Recall : Let H and K be finite subgroups of a group G.
Then HK is a subgroup of G if and only if HK = KH
and O(HK) = O(H)O(K) / O(H∩K)
Let H and K be subgroups of a group G.
If one of H and K is normal in G then
HK is a subgroup of G.
If both H and K are normal in G then
HK is a normal subgroup of G.
If p is an odd prime then Dp and ℤ2p are the only groups of order 2p.
Proof Let G be a group of order 2p, where p is an odd prime.
By Cauchy’s theorem G has a subgroup H of order 2 and a subgroup
K of order p.
Let H = <a> = {ai / i = 0, 1} and K = <b> = {bj / j = 0, . . . , p-1}.
Since O(G) / O(K) = 2, index of K is 2, K is a normal subgroup of G.
Hence HK is a subgroup of G.
If x ∈ H ∩ K then O(x) | 2 and O(x) | p. Hence O(x) = 1 and hence H ∩ K = {e} and O(HK) = O(H) O(K) = 2p = O(G).
Therefore G = HK = {ai bj / i = 0, 1, j = 0, . . . , p-1}.
Since K is normal in G, aba-1 ∈ K. i.e. aba-1 = bk for some 1 ≤ k ≤ p-1.
Hence bkk = (aba-1)k = abka-1 = a(aba-1)a-1 = a2ba-2 = b. i.e. bkk - 1 = e.
Therefore p | k2-1= (k + 1)(k - 1). So, either p | (k + 1) or p | (k - 1).
Since 1 ≤ k ≤ p - 1, either k + 1 = p or k - 1 = 0.
i.e. either k = p - 1 or k =1.
Hence either aba-1 = bp-1 or aba-1 = b and hence
either ab = bp-1a or ab = ba .
If ab = bp-1a , G = Dp . If ab = ba then O(ab) = O(a)O(b) = 2p = O(G),
hence G is cyclic and G = ℤ2p. Therefore either G = Dp or ℤ2p.
In general, suppose p and q are primes with p < q.
i) If p ∤ q - 1, then ℤpq is the only one group of order pq.
ii) If p | q - 1, then there exists a unique
non-abelian group and an abelian group of
order pq.
G = {aibj / 0 ≤ i < p and 0 ≤ j < q},
where ap = e = bq and aba-1 = br,
for some 1 < r < q and p is the smallest
positive integer such that rp ≡ 1(mod q)
and ℤpq are the only groups of order pq.
i.e. there are only two groups of order pq.
For proof visit https://velrajanm.blogspot.com/2024/04/groups-of-order-pq.html
ℤn is the only group of order n if and only if n and φ(n) are relatively prime.
Let H and K be groups . Then H × K is a group under the operation
(a, b)(c, d) = (ac, bd). And O(H × K) = O(H)O(K).
Also, H × K is isomorphic to K × H.
If H1, . . . , Hn are groups then H1 × . . . × Hn is a group and it is called the direct product of H1, . . . , Hn .
H1 × . . . × Hn is abelian if and only if H1, . . . , Hn are abelian
Learning from this result : We can get an infinite number of groups from a known one, since, if H is a group then H × H × . . . × H is a group.
The order of (a,b) in H × K is the l.c.m of the orders of a and b.
Suppose H and K are cyclic groups of order m and n respectively. Then H × K is
cyclic if and only if m and n are relatively prime.
For Cyclic Groups Lecture Notes visit
https://drive.google.com/file/d/1syDs_S3UMSqD8Q0_3NdbA1B0HhOwvxll/view?usp=drivesdk
Suppose H1, . . . , Hn are cyclic groups of orders m1, . . . , mn. Then H1 × . . . × Hn is a cyclic group if and only if m1, . . . , mn are pairwise relatively prime.
If H1, . . . , Hn are cyclic groups then H1 × . . . × Hn is an abelian group.
Learning from this result : We can get examples of groups of any given order. For example,
ℤ4 × â„¤9 ≅ ℤ36, since (4, 9) = 1, ℤ4 × â„¤9 is a cyclic group of order 4 × 9 = 36 it is ≅ ℤ36.
ℤ4 × â„¤9 × â„¤15 is an abelian group of order 4 × 9 × 15 = 540, but it is not cyclic, since (9, 15) = 3.
ℤ4 × S5 × â„¤10 is a non-abelian group of order 4 × 5! × 10 = 4800.
ℤ4 × D60 × â„¤10 is a non-abelian group of order 4800.
These two groups of order 4800 are non isomorphic.
D60 has an element of order 60, but S5 has no elements of order 60.
Learning from the above and earlier results : We can get isomorphic and non-isomorphic abelian groups of any given order from ℤn s.
For example, we find abelian groups of order 180.
First note that 180 = 22 × 32 × 5. Think of different ways of expressing factorization of 180.
180 = 22 × 32 × 5 = 2 × 2 × 9 × 5 = 2 × 18 × 5 = 2 × 9 × 10 = 4 × 3 × 3 × 5
= 36 × 5 = 4 × 45 = 20 × 9 = . . . .
For each factorization of 180, we can get an abelian group of order 180.
Yes, corresponding to 22 × 32 × 5 we get ℤ4 × â„¤9 × â„¤5 , for 2 × 2 × 9 × 5 we get ℤ2 × â„¤2 × â„¤9 × â„¤5 and so on.
Then think how to determine isomorphic groups among them.
For any n, ℤn is cyclic, so to get isomorphic cyclic groups of order n we have to consider all the different factorization of n as pairwise relatively prime factors.
4 × 9 × 5; 36 × 5; 4 × 45; 20 × 9 are the only factorizations of 180 in which the factors are pairwise relatively prime. Hence ℤ4 × â„¤9 × â„¤5 , ℤ36 × â„¤5 , ℤ4 × â„¤45 , ℤ20 × â„¤9 are cyclic groups of order 180 and hence they are isomorphic to ℤ180.
Since (9, 5) = 1 = (2, 9) = (2, 5) = (10, 9), we get
ℤ9 × â„¤5 ≅ ℤ45, ℤ2 × â„¤9 ≅ ℤ18, ℤ2 × â„¤5 ≅ ℤ10, ℤ10 × â„¤9 ≅ ℤ90. Hence
ℤ2 × â„¤2 × â„¤9 × â„¤5 ≅ ℤ2 × â„¤2× â„¤45 ≅ ℤ2 × â„¤18 × â„¤5
≅ ℤ10 × â„¤18 ≅ ℤ2 × â„¤10 × â„¤9 ≅ ℤ2 × â„¤90 are abelian but not cyclic groups of order 180.
Since 2, 2 are not relatively prime ℤ2 × â„¤2 is not cyclic hence
ℤ2 × â„¤2 × â„¤9 × â„¤5 is not cyclic and hence ℤ2 × â„¤2 × â„¤9 × â„¤5 is not isomorphic to ℤ180.
In this way we can verify that
i) ℤ180 ≅ ℤ4 × â„¤9 × â„¤5 ≅ ℤ36 × â„¤5 ≅ ℤ4 × â„¤45 ≅ ℤ20 × â„¤9
ii) ℤ4 × â„¤3 × â„¤3 × â„¤5 ≅ ℤ12 × â„¤3 × â„¤5 ≅ ℤ20 × â„¤3 × â„¤3
≅ ℤ4 × â„¤3 × â„¤15 ≅ ℤ12 × â„¤15
iii) ℤ2 × â„¤2 × â„¤9 × â„¤5 ≅ ℤ2 × â„¤2× â„¤45 ≅ ℤ2 × â„¤18 × â„¤5
≅ ℤ10 × â„¤18 ≅ ℤ2 × â„¤10 × â„¤9 ≅ ℤ2 × â„¤90
iv) ℤ2 × â„¤2 × â„¤3 × â„¤3 × â„¤5 ≅ ℤ2 × â„¤2 × â„¤3 × â„¤15 ≅ ℤ2 × â„¤3 × â„¤3 × â„¤10
≅ ℤ2 × â„¤3 × â„¤6 × â„¤5
are non-isomorphic abelian groups of order 180.
LCM {4, 3, 3, 5} = 60, so the order of any (a, b, c, d) in ℤ4 × â„¤3 × â„¤3 × â„¤5 is at most 60. But in ℤ2 × â„¤2 × â„¤9 × â„¤5 the order of (1, 1, 1, 1) = LCM {2, 2, 9, 5} = 90. Hence ℤ2 × â„¤2 × â„¤9 × â„¤5 is not isomorphic to ℤ4 × â„¤3 × â„¤3 × â„¤5. Similarly justify the other claims.
And the above 4 groups are the only non-isomorphic abelian groups of order 180.
(This follows from Fundamental theorem of finite abelian groups :
Any finite abelian group is a direct product of cyclic groups of prime-power orders.)
Any group of order p2 is abelian, where p is prime.
Learning from this result : p2 = p × p. Hence ℤp2 and ℤp × â„¤p are the
only groups of order p2. ℤp × â„¤p is abelian but not cyclic ?
Oh! Just by the number n - the order of the group and its factorization we are able to determine the groups of the given order.
In the following cases by the order of the group we are able to determine the groups of the given order.
1. For any prime p, (ℤp, ⊕p) is the only group of order p.
And ℤp2 and ℤp × â„¤p are the only groups of order p2.
2. If p is an odd prime then Dp and ℤ2p are the only groups of order 2p.
3. Suppose p and q are primes with p < q.
i) If p ∤ q - 1, then ℤpq is the only one group of order pq.
ii) If p | q - 1, then the unique non-abelian group
G = {aibj / 0 ≤ i < p and 0 ≤ j < q},
where ap = e = bq and aba-1 = br,
for some 1 < r < q and p is the smallest
positive integer such that rp ≡ 1(mod q)
and the unique abelian group ℤpq
are the only groups of order pq.
4. ℤn is the only group of order n if and only if n and φ(n) are relatively prime
In the other cases by the factorization of order of the group we are able to determine the groups of the given order.
What wonderful relations between groups and numbers.
For n ≤ 8 we find all groups (up to isomorphism) of order n.
n = 1 {e} = ℤ1
n = 2 {e, a} ≅ ℤ2, since a ≠ e, a2 ≠ a, hence a2 = e
n = 3 prime, cyclic ≅ ℤ3
n = 4 if cyclic ≅ ℤ4 . Suppose G is a non-cyclic group of order 4.
For any x ≠ e ∈ G, since O(x) | O(G) = 4, O(x) = 2. Hence x2 = e , for all x ∈ G and hence G is abelian and G ≅ ℤ2 × â„¤2 ≅ V4.
Learning :
The Klein 4 group V4 = {e, a, b, c}, where a2 = b2 = c2 = e and ab = c.
Hence (e) = {e}, (a) = {e, a}, (b) = {e, b}, (c) = {e, c}. And hence no element generates V4 i.e. V4 is not cyclic.
Since O(V4) = 4, by Lagrange's theorem, the order of any subgroup of V4 is 1 or 2 or 4.
Hence (e)= {e}, (a) = {e, a}, (b) = {e, b}, (c) = {e, c} and V4 are the only subgroups of V4.
We note that all the proper subgroups of V4 are cyclic, but V4 is not cyclic.
Since 2, 3 are prime, any group of order 2 or 3 is cyclic. And V4 is a group of order 4. Hence V4 is the smallest group which is not cyclic but all the proper subgroups are cyclic.
n = 5 prime, cyclic ≅ ℤ5
n = 6
Suppose G is a group of order 6. Since O(G) is even, there is an b ≠ e ∈ G
with b2 = e. Suppose x2 = e for all x ∈ G, then G is abelian and for x ≠ e, b ∈ G,
{e, b, x, xb} is a subgroup of order 4, a contradiction.
Hence there is an a ∈ G s.t. a2 ≠ e. Hence
O(a) = 6 or 3. If O(a) = 6 then G is cyclic and G ≅ ℤ6.
If O(a) = 3, let H = <a> = {e, a, a2}.
Since the index of H is 2, H is normal. Clearly, b not in H , hence G = H ∪ Hb.
i.e. G = {e, a, a2, b, ab, a2b}. Since H is normal, bab-1∈ H.
If bab-1 = e then a = e. Hence
bab-1 = a or a2 and hence ba = ab or ba = a2b.
If ba = ab then O(ab) = O(a)O(b) = 6, and G = <ab> ≅ ℤ6.
If ba = a2b then G ≅ D3 ≅ S3.
Learning :
In S3 = {e, (1 2), (1 3), (2 3), (1 2 3), (1 3 2) }, (1 2)2 = e = (1 3)2 = (2 3 )2 ,
(1 2 3)2 = (1 3 2), (1 2 3)3 = e, (1 3 2)2 = (1 2 3), (1 3 2)3 = e. Hence
<e> = {e}, <(1 2)> = {e, (1 2)}, <(1 3)> = {e, (1 3)}, <(2 3)> = {e, (2 3)},
<(1 2 3)> = {e, (1 2 3), (1 3 2)} = <(1 3 2)>.
Hence no element generates S3 i.e. S3 is not cyclic.
Since O(S3) = 6, by the Lagrange's theorem, the order of any subgroup of S3 is 1 or 2 or 3 or 6.
Hence <e> = {e}, <(1 2)> = {e, (1 2)}, <(1 3)> = {e, (1 3)}, <(2 3)> = {e, (2 3)},
<(1 2 3)> = {e, (1 2 3), (1 3 2)} = <(1 3 2)> and S3 are the only subgroups of S3.
We note that all the proper subgroups of S3 are cyclic, but S3 is not cyclic. Also
all the proper subgroups of S3 are abelian, but S3 is non-abelian. And S3 is the smallest non-abelian group(all the groups of order < 6 are abelian, but S3 is the only non-abelian group of order 6.
n = 7 prime, cyclic ≅ ℤ7
n = 8 Suppose G is a group of order 8.
i) Suppose G is abelian then it is isomorphic to one of ℤ8 , ℤ4 × â„¤2 , ℤ2 × â„¤2 × â„¤2 .
8 = 4 × 2 = 2 × 2 × 2, LCM{4, 2} = 4, LCM{2, 2, 2} = 2.
ii) Suppose G is non-abelian. Suppose x2 = e for all x then G is abelian, hence there is an a ∈ G with a2 ≠ e. Since O(a) | 8 and G is not cyclic O(a) = 4.
Let H = <a> = {e, a, a2 , a3}. Since the index of H is 2, H is normal.
Let b not in H , then G = H ∪ Hb and H ∩ Hb = Φ.
i.e. G ={e, a, a2, a3, b, ab, a2b, a3b}. Since H is normal, bab-1 ∈ H.
If bab-1 = e then a = e.
If bab-1 = a then ba = ab and hence G is abelian.
If bab-1 = a2 then (bab-1)2 = a4 = e
i.e. ba2b-1 = e i.e. a2 = e. Therefore bab-1 = a3 i.e. ba = a3b.
Since b is not in H, b2 is not in Hb. Hence b2 ∈ H. If b2 = a or a3 then O(b) = 8, and G = <b>. Hence b2 = e or a2.
Therefore G = {e, a, a2, a3, b, ab, a2b, a3b}, where a4 = e, ba = a3b and b2 = e or a2 . Hence G ≅ D4 or Q 4.
Therefore, ℤ8 , ℤ4 × â„¤2 , ℤ2 × â„¤2 × â„¤2 and D4 are the only groups of order 8.
A group is simple if it's only normal subgroups are the identity subgroup and the
group itself.
The abelian simple groups are precisely ℤn where n = 1 or n is prime.
A5 is the only non-abelian simple group whose order is less than 168.
An is simple for n ≥ 5.
There are only 5 non-abelian simple groups of order less than 1000 and only 56 of order less than 10,00,000.
In January 1981 Gorenstien announced that, complete list of all finite simple groups had been found. The proof that this list is complete runs over 10000 journal pages.
One more relation between numbers and groups
Let n = 9 and 1 ≤ s < n. What is the smallest positive integer m such that ms is a multiple of n ?
We note that m = n/d , where d = (n, s).
In general, if n is any positive integer and 1 ≤ s < n then n/d is the smallest positive integer such that (n/d)s is a multiple of n.
For, Suppose d = (n, s). Then (n/d) s = n (s/d), is a multiple of n and 1 < n/d ≤ n.
Let m be a positive integer such that ms is a multiple of n and ms = nq for some positive integer q.
Then (ms)/d = (nq)/d i.e. m(s/d) = (n/d)q.
Since (s/d , n/d) = 1, (n/d) | m.
Hence nd is the smallest positive integer such that (nd)s is a multiple of n.
We can see that this result on numbers is the base of the following theorem on groups.
Theorem Let G be a group, a ∈ G and the order of a = n. Then for any s ≥ 1, the order of as = n/d , where d = (n, s), g.c.d. of n and s.
Proof Let d = (n, s).
Then (as) n/d = as(n/d) = an(s/d) = (an)s/d = es/d = e (since s(n/d) = n(s/d) and an = e).
Suppose (as)m =asm = e. Then, since the order of a = n, n | sm.
i.e. sm = nx for some integer x.
Hence (s/d )m = (n/d)x and hence (n/d) | (s/d)m.
Since d = (n, s), (n/d, s/d) = 1. Hence (n/d) | m.
Therefore, the order of as = n/d .
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