Cauchy's Theorem (group Theory)
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Cauchy's Theorem Group Theory

In mathematics, specifically group theory, Cauchy's theorem states that if G is a finite group and p is a prime number dividing the order of G (the number of elements in G), then G contains an element of order p. That is, there is x in G such that p is the smallest positive integer with xp = e, where e is the identity element of G. It is named after Augustin-Louis Cauchy, who discovered it in 1845.[1][2]

The theorem is related to Lagrange's theorem, which states that the order of any subgroup of a finite group G divides the order of G. Cauchy's theorem implies that for any prime divisor p of the order of G, there is a subgroup of G whose order is p--the cyclic group generated by the element in Cauchy's theorem.

Cauchy's theorem is generalised by Sylow's first theorem, which implies that if pn is the maximal power of p dividing the order of G, then G has a subgroup of order pn (and using the fact that a p-group is solvable, one can show that G has subgroups of order pr for any r less than or equal to n).

## Statement and proof

Many texts prove the theorem with the use of strong induction and the class equation, though considerably less machinery is required to prove the theorem in the abelian case. One can also invoke group actions for the proof.[3]

Cauchy's theorem — Let G be a finite group and p be a prime. If p divides the order of G, then G has an element of order p.

### Proof 1

We first prove the special case that where G is abelian, and then the general case; both proofs are by induction on n = |G|, and have as starting case n = p which is trivial because any non-identity element now has order p. Suppose first that G is abelian. Take any non-identity element a, and let H be the cyclic group it generates. If p divides |H|, then a|H|/p is an element of order p. If p does not divide |H|, then it divides the order [G:H] of the quotient group G/H, which therefore contains an element of order p by the inductive hypothesis. That element is a class xH for some x in G, and if m is the order of x in G, then xm = e in G gives (xH)m = eH in G/H, so p divides m; as before xm/p is now an element of order p in G, completing the proof for the abelian case.

In the general case, let Z be the center of G, which is an abelian subgroup. If p divides |Z|, then Z contains an element of order p by the case of abelian groups, and this element works for G as well. So we may assume that p does not divide the order of Z; since it does divide |G|, there is at least one conjugacy class of a non-central element a whose size is not divisible by p. But the class equation shows that size is [G : CG(a)], so p divides the order of the centralizer CG(a) of a in G, which is a proper subgroup because a is not central. This subgroup contains an element of order p by the inductive hypothesis, and we are done.

### Proof 2

This proof uses the fact that for any action of a (cyclic) group of prime order p, the only possible orbit sizes are 1 and p, which is immediate from the orbit stabilizer theorem.

The set that our cyclic group shall act on is the set

${\displaystyle X=\{\,(x_{1},\ldots ,x_{p})\in G^{p}:x_{1}x_{2}\cdots x_{p}=e\,\}}$

of p-tuples of elements of G whose product (in order) gives the identity. Such a p-tuple is uniquely determined by all its components except the last one, as the last element must be the inverse of the product of those preceding elements. One also sees that those elements can be chosen freely, so X has |G|p-1 elements, which is divisible by p.

Now from the fact that in a group if ab = e then also ba = e, it follows that any cyclic permutation of the components of an element of X again gives an element of X. Therefore one can define an action of the cyclic group Cp of order p on X by cyclic permutations of components, in other words in which a chosen generator of Cp sends

${\displaystyle (x_{1},x_{2},\ldots ,x_{p})\mapsto (x_{2},\ldots ,x_{p},x_{1})}$.

As remarked, orbits in X under this action either have size 1 or size p. The former happens precisely for those tuples ${\displaystyle (x,x,\ldots ,x)}$ for which ${\displaystyle x^{p}=e}$. Counting the elements of X by orbits, and reducing modulo p, one sees that the number of elements satisfying ${\displaystyle x^{p}=e}$ is divisible by p. But x = e is one such element, so there must be at least other solutions for x, and these solutions are elements of order p. This completes the proof.

## Uses

A practically immediate consequence of Cauchy's theorem is a useful characterization of finite p-groups, where p is a prime. In particular, a finite group G is a p-group (i.e. all of its elements have order pk for some natural number k) if and only if G has order pn for some natural number n. One may use the abelian case of Cauchy's Theorem in an inductive proof[4] of the first of Sylow's theorems, similar to the first proof above, although there are also proofs that avoid doing this special case separately.

### Example1

Let G is a finite group where x2 = e for all element x of G. Then G has the order 2n for some non negative integer n. Let |G| is m. In the case of m is 1, then G = {e}. In the case of m ≥ 2, if m has the odd prime factor p, G has the element x where xp = e from Cauchy's theorem. It conflicts with the assumption. Therefore m must be 2n.[5] The well-known example is Klein four-group.

### Example2

An Abelian simple group is either {e} or cyclic group Cp whose order is a prime number p. Let G is an Abelian group, then all subgroups of G are normal subgroups. So, if G is a simple group, G has only normal subgroup that is either {e} or G. If |G| = 1, then G is {e}. It is suitable. If |G| ≥ 2, let aG is not e, the cyclic group a is subgroup of G and a is not {e}, then G = ⟨a. Let n is the order of a. If n is infinite, then

${\displaystyle \{e\}\subsetneqq \langle a^{2}\rangle \subsetneqq \langle a\rangle =G.}$

So in this case, it is not suitable. Then n is finite. If n is composite, n is divisible by prime q which is less than n. From Cauchy's theorem, the subgroup H will be exist whose order is q, it is not suitable. Therefore, n must be a prime number.

## Notes

1. ^
2. ^
3. ^
4. ^ Jacobson 2009, p. 80.
5. ^ Finite groups where x2=e has order 2n, Stack Exchange, 2015-09-23