Cauchy's Group Theorem

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This proof is about Cauchy's Group Theorem. For other uses, see Cauchy's Theorem.

Theorem

Let $G$ be a finite group whose identity is $e$.

Let $p$ be a prime number which divides the order of $G$.


Then $G$ has a subgroup of order $p$.


Proof 1

Lemma

Let $G$ be a finite abelian group whose identity is $e$.

Let $p$ be a prime number which divides the order of $G$.


Then $G$ has a subgroup of order $p$.


Proof of Lemma

Let $\order G$ be a prime number.

Then from Prime Group is Cyclic $G$ has a generator $\gen g$ such that $\order g = p$.


Now suppose $e \ne g \in G$.

Let $\order g = n$.

Let $p \divides n$.

Then by Subgroup of Finite Cyclic Group is Determined by Order the cyclic group $\gen g$ has an element $g$ of order $p$.


Suppose $p \nmid n$.

From Subgroup of Abelian Group is Normal, $\gen g$ is normal in $G$.

Consider the quotient group $G' = \dfrac G {\gen g}$.

As $p \nmid n$, $\gen e \subsetneq \gen g \subsetneq G$.

Thus $\order {G'} < \order G$.

But we have that $p \divides \order G$.


It follows by induction that $G'$ has an element $h'$ of order $p$.

Let $h$ be a preimage of $h'$ under the quotient epimorphism $\phi: G \to G'$.

Then:

$\paren {h'}^p = e'$

where $e'$ is the identity of $G'$.

Thus $h^p \in \gen g$ so $\paren {h^p}^n = \paren {h^n}^p = e$.

Thus either $h^n$ has order $p$ or $h^n = e$.

If $h^n = e$ then $\paren {h'}^n = e'$.

But since $p$ is the order of $h'$ it would follow that $p \divides n$, contrary to assumption.

$\Box$


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We prove the theorem by induction on the order $\order G$.

The first few cases $\order G = 1,2,3$ are obvious. For the induction step, start with the class equation

$\order G = 1 + \order {C_2} + \cdots + \order {C_r}$

Since $p \divides \order G$, we must have $p \nmid \order {C_j}$ for some $j \geq 2$,

where $C_j$ are conjugacy classes in $G \setminus \set e$.

Let $x \in C_j$,

it follows that $p \divides \order {\map {C_G} x}$,

where $\map {C_G} x$ is the centralizer of $x$ in $G$,

since $\order {C_j} = \frac{\order G}{\order {\map {C_G} x}}$ by Number of Elements in Conjugacy Class Equals the Index of Centralizer.

If $\map {C_G} x \neq G$, then by induction $\map {C_G} x$ contains an element of order $p$, and this element also belongs to $G$.

Otherwise, $\map {C_G} x = G$, which implies that $x \in \map Z G$ (the center of $G$), and by choice $x \neq e$, so $\map Z G \neq \set e$.

Either $p \divides \order {\map Z G}$ or $p \nmid \order {\map Z G}$.

In the first case the proof reduces to lemma (the abelian case).

In the second case, by induction hypothesis to the quotient group $G / \map Z G$, there exists $x \in G$ such that the image $\bar{x} \in G / \map Z G$ has order $p$,

That is, $x^p \in \map Z G$ but $x \notin \map Z G$.

Let $X$ be the cyclic group generated by $x$.

Now the product $X \map Z G$ is abelian and has order divisible by $p$, so by lemma (the abelian case) it has an element of order $p$, and again this element also belongs to $G$.

This completes the induction step, and with it the proof.

$\blacksquare$


Proof 2

This result follows as a special case of Group has Subgroups of All Prime Power Factors.

$\blacksquare$


Also known as

Cauchy's Group Theorem is also seen referred to as Cauchy's Theorem, but as there are several theorems named so, this is not endorsed by $\mathsf{Pr} \infty \mathsf{fWiki}$.


Source of Name

This entry was named for Augustin Louis Cauchy.


Historical Note

Cauchy's Group Theorem was proved by Augustin Louis Cauchy in $1845$.


Sources

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