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