cauchy theorem proof
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Answer:
Step-by-step explanation:
PROOF OF CAUCHY’S THEOREM
KEITH CONRAD
The converse of Lagrange’s theorem is false in general: if G is a finite group and d | |G|
then G doesn’t have to contain a subgroup of order d. (For example,|A4| = 12 and A4
has no subgroup of order 6). We will show the converse is true when d is prime. This is
Cauchy’s theorem.
Theorem. (Cauchy 1845) Let G be a finite group and p be a prime factor of |G|. Then
G contains an element of order p. Equivalently, G contains a subgroup of order p.
The equivalence of the existence of an element of order p and a subgroup of order p is
easy: an element of order p generates a subgroup of order p, while conversely any nonidentity
element of a subgroup of order p has order p because p is prime.
Before treating Cauchy’s theorem, let’s see that the special case for p = 2 can be proved
in a simple way. If |G| is even, consider the set of pairs {g, g−1}, where g 6= g
−1
. This
takes into account an even number of elements of G. Those g’s that are not part of such
a pair are the ones satisfying g = g
−1
, i.e., g
2 = e. Therefore if we count |G| mod 2, we
can ignore the pairs {g, g−1} where g 6= g
−1 and we obtain |G| ≡ |{g ∈ G : g
2 = e}| mod 2.
One solution to g
2 = e is e. If it were the only solution, then |G| ≡ 1 mod 2, which is false.
Therefore some g0 6= e satisfies g
2
0 = e, which gives us an element of order 2.
Now we prove Cauchy’s theorem.
Proof. We will use induction on |G|.
1 Let n = |G|. Since p | n, n ≥ p. The base case is
n = p. When |G| = p, any nonidentity element of G has order p because p is prime. Now
suppose n > p, p | n, and the theorem is true for all groups having order less than n that is
divisible by p. We will treat separately abelian G (using homomorphisms) and nonabelian
G (using conjugacy classes).
Case 1: G is abelian.
Assume no element of G has order p and we will get a contradiction.
No element has order divisible by p: if g ∈ G has order r and p | r then g
r/p would have
order p.
Let G = {g1, g2, . . . , gn} and let gi have order mi
, so each mi
is not divisible by p. Let m
be the least common multiple of the mi
’s, so m is not divisible by p and g
m
i = e for all i.
Because G is abelian, the function f : (Z/(m))n → G given by f(a1, . . . , an) = g
a1
1
· · · g
an
n
is
a homomorphism:
2
f(a1, . . . , an)f(b1, . . . , bn) = f(a1 + b1, . . . , an + bn).
That is,
g
a1
1
· · · g
an
n
g
b1
1
· · · g
bn
n = g
a1
1
g
b1
1
· · · g
an
n
g
bn
n = g
a1+b1
1
· · · g
an+bn
n
1Proving a theorem on groups by induction on the order of the group is a very fruitful idea in group
theory.
2This function is well-defined because g
m
i = e for all i, so g
a+mk
i = g
a
i
for any k ∈ Z.
1
2 KEITH CONRAD
from commutativity of the gi
’s. This homomorphism is surjective (each element of G is a
gi
, and if ai = 1 and other aj ’s are 0 then f(a1, . . . , an) = gi), so by the first isomorphism
theorem (Z/(m))n/ ker f ∼= G. Therefore
|G| =
|(Z/(m))n
|
| ker f|
=
mn
| ker f|
,
so |G|| ker f| = mn
. Thus |G| is a factor of mn
, but p divides |G| and mn
is not divisible by
p, so we have a contradiction.
Case 2: G is nonabelian.
Assume no element of G has order p and we will get a contradiction.
In every proper subgroup H of G there is no element of order p (H may be abelian or
nonabelian), so by induction no proper subgroup of G has order divisible by p. For each
proper subgroup H, |G| = |H|[G : H] and |H| is not divisible by p while |G| is divisible by
p, so p | [G : H] for every proper subgroup H of G.
Since G is nonabelian it has some conjugacy classes with size greater than 1. Let these
be represented by g1, g2, . . . , gk. Conjugacy classes in G of size 1 are the elements in Z(G).
Since the conjugacy classes in G form a partition of G, computing |G| by adding the sizes
of its conjugacy classes implies