Cauchy's theorem (group theory)

Cauchy's theorem is a theorem in the mathematics of group theory, named after Augustin Louis Cauchy. It 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" so that "p" is the lowest non-zero number with "x"^"p" = "e", where "e" is the identity element.

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 "p"^"n" is any prime power dividing the order of "G", then G has a subgroup of order "p"^"n".

tatement and proof

Many texts appear to 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.

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 induct on "n" = |"G"| and consider the two cases where "G" is abelian or "G" is nonabelian. Suppose "G" is abelian. If "G" is simple, then it must be cyclic of prime order and trivially contains an element of order "p". Otherwise, there exists a nontrivial, proper normal subgroup $H\; riangleleft\; G$. If "p" divides |"H"|, then "H" contains an element of order "p" by the inductive hypothesis, and thus "G" does as well. Otherwise, "p" must divide the index ["G":"H"] by Lagrange's theorem, and we see the quotient group "G"/"H" contains an element of order "p" by the inductive hypothesis; that is, there exists an "x" in "G" such that ("Hx")^"p" = "Hx"^"p" = "H". Then there exists an element "h"_"1" in "H" such that "h"_"1""x"^"p" = 1, the identity element of "G". It is easily checked that for every element "a" in "H" there exists "b" in "H" such that "b"^"p" = "a," so there exists "h"_"2" in "H" so that "h"_"2" ^"p" = "h"_"1". Thus "h"_{"2""x" has order "p", and the proof is finished for the abelian case.}

Suppose that "G" is nonabelian, so that its center "Z" is a proper subgroup. If "p" divides the order of the centralizer "C"_"G"("a") for some noncentral element "a" (i.e. "a" is not in "Z"), then "C"_"G"("a") is a proper subgroup and hence contains an element of order "p" by the inductive hypothesis. Otherwise, we must have "p" dividing the index ["G":"C"_"G"(a)] , again by Lagrange's Theorem, for all noncentral "a". Using the class equation, we have "p" dividing the left side of the equation (|"G"|) and also dividing all of the summands on the right, except for possibly |"Z"|. However, simple arithmetic shows "p" must also divide the order of "Z", and thus the center contains an element of order "p" by the inductive hypothesis as it is a proper subgroup and hence of order strictly less than that of "G". This completes the proof.

Proof 2: This time we define the set of p-tuples whose elements are in the group G by $X\; =\; left\; \{\; mathbf\{x\}\; =\; (x\_1,cdots,x\_p)\; in\; G\; imes\; cdots\; imes\; G\; :\; prod\_\{i=1\}^p\; x\_i\; =\; 1\; ight\; \}$

Note that we can choose only (p-1) of the $x\_i$ independently, since we are constrained by the product equal to 1. Thus $left\; |\; X\; ight\; |\; =\; left\; |\; G\; ight\; |^\{p-1\}$, from which we deduce that "p" also divides $left\; |X\; ight\; |$

Define the action $ho\; :\; mathbb\{Z\}\_p\; imes\; X\; o\; X\; ,,$ by $ho\; (z^r,mathbf\{x\})\; =\; (x\_\{r+1\},cdots,x\_p,\; x\_1,cdots\; x\_r)$ , where $mathbb\{Z\}\_p$ is the cyclic group of order "p"
Then $O(\; mathbf\{x\})\; =\; left\; \{\; z^r\; mathbf\{x\}\; in\; X\; :\; z\; in\; mathbb\{Z\}\_p\; ,\; r\; in\; mathbb\{Z\}\; ight\; \}\; ,$ is the orbit of some element $mathbf\{x\}\; in\; X$
The stabilizer is $S(mathbf\{x\}\; )=\; left\; \{\; z\; in\; mathbb\{Z\}\_p\; :\; z\; mathbf\{x\}\; =\; mathbf\{x\}\; ight\; \}\; ,$, from which we can deduce the order,

We have from the Orbit-Stabilizer Theorem that $left\; |\; mathbb\{Z\}\_p\; ight\; |=\; left\; |\; S(mathbf\{x\})\; ight\; |\; left\; |O(mathbf\{x\})\; ight\; |$ for each $mathbf\{x\}\; in\; X$
Take $mathbf\{x\}\_1\; =\; (1,cdots,1)$ and $O(mathbf\{x\}\_j)$ the distinct orbits. Then $left\; |\; O(mathbf\{x\}\_1)\; ight\; |\; =\; frac\{mathbb\{Z\}\_p\}\{left\; |S(mathbf\{x\}\_1)\; ight\; =\; frac\{p\}\{p\}\; =\; 1$
Hence we know that $left\; |X\; ight\; |\; =\; left\; |O(mathbf\{x\}\_1)\; ight\; |\; +\; sum\_\{j=2\}^m\; left\; |O(mathbf\{x\}\_j)\; ight\; |=\; 1+\; sum\_\{j=2\}^m\; left\; |O(mathbf\{x\}\_j)\; ight\; |$
"p" divides |X| implies that there is at least one other $mathbf\{x\}\_j$ with the property that its orbit has order 1
Then we have $mathbf\{x\}\_j\; =\; (x\_j,cdots,x\_j)\; Rightarrow\; overbrace\{x\_j\; cdots\; x\_j\}^p\; =\; x\_j^p\; =\; 1$ by the definition of "X"
Since "x_j" is in "G" 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 "p"^"k" for some natural number "k") if and only if "G" has order "p"^"n" for some natural number "n".

References

* James McKay. "Another proof of Cauchy's group theorem", American Math. Monthly, 66 (1959), p. 119.

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