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# A jargon-minimal counting proof of Sylow’s first theorem

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October 2023 (Vol. 55, No. 5)

#### Abstract

We present a short proof of Sylow’s famous ‘First Theorem.’  Stripped to its essentials, the proof—attributed to Wielandt  (1959)—relies only on basic concepts (equivalence relations and divisibility). Whereas most textbook proofs invoke plenty of group-theoretic jargon (stabilizers, conjugacy  classes,  etc.),  this  one  avoids  all  that,  stands  alone,  and fits on a page.

A (binary) million years ago, Helmut Wielandt published an appealing counting proof [3] of Sylow’s First Theorem.  Stripped to its bare essentials, it requires surprisingly little group theory and ought to be better known. The version below lends itself to an hour or so in a classroom with second- or third-year university math enthusiasts.

Theorem (‘Sylow’s First’) If $p$ is a prime, $r$ is an integer coprime to $p$, and $G$ is a group of order $p^{\alpha}r$, then $G$ contains a(t least one) subgroup of order $p^{\alpha}$.

Proof. Let $\mathcal{S}$ denote the family of all subsets of $G$ of size $p^{\alpha}$, and write $\mathcal{S}=\{A_1,A_2,\ldots,A_n\}$. We shall argue that at least one member of $\mathcal{S}$ is a group—indeed, is a subgroup of $G$.

It’s convenient to know that $p$ and $n$ are coprime, so we first establish this fact. We have

$n=\binom{p^{\alpha}r}{p^{\alpha}}=\frac{p^{\alpha}r(p^{\alpha}r-1)(p^{\alpha}r-2)\cdots(p^{\alpha}r-(p^{\alpha}-1))}{p^{\alpha}\cdot1\cdot 2\cdot~\cdots~\cdot(p^{\alpha}-1)}=r\prod\limits_{k=1}^{p^{\alpha}-1}\frac{p^{\alpha}r-k}{k}.$

Consider the factors $\frac{p^{\alpha}r-k}{k}$. If $p\nmid k$, then $p\nmid (p^{\alpha}r-k)$. On the other hand, if $p\,|\, k$, then $k=p^{\beta}s$ for some integer $\beta$ with $1\leq\beta<\alpha$ and $p\nmid s$. So here, we have

$\frac{p^{\alpha}r-k}{k}=\frac{p^{\beta}(p^{\alpha-\beta}r-s)}{p^{\beta}s}=\frac{p^{\alpha-\beta}r-s}{s},$

and $p\nmid p^{\alpha-\beta}r-s$ (for otherwise, $p\,|\, s$). In either case, $p$ fails to divide each factor $\frac{p^{\alpha}r-k}{k}$, and thus we see that $p\nmid n$.

Now if $x\in G$ and $A_i\in\mathcal{S}$, then the set $A_{i}x$ also contains $p^{\alpha}$ elements (for the map $A_i\to A_{i}x$ given by $a\mapsto ax$ is injective). Hence, $A_{i}x=A_{j}$ for some $j\in\{1,\ldots,n\}$. Let us define a relation $\sim$ on $\mathcal{S}$ by

$A_i\sim A_j \Longleftrightarrow A_{i}x=A_{j} \text{ for some } x\in G.$

Using the group axioms for $G$, it’s easy to see that $\sim$ is an equivalence relation. Since $p\nmid n$, at least one equivalence class $\mathcal{C}$  of $\sim$ contains some $q$ sets $A_i$ with $p\nmid q$; say $\mathcal{C}=\{A_1,A_2,\ldots,A_q\}$ (relabelling the $A_i$‘s as necessary).

Let $H=\{x\in G\colon A_{1}x=A_{1}\}$; one easily verifies that $H$ is a subgroup of $G$—write $h$ for its order. For $x,y\in G$, observe that

$A_{1}x=A_{1}y \Leftrightarrow A_{1}xy^{-1}=A_{1} \Leftrightarrow xy^{-1}\in H \Leftrightarrow Hxy^{-1}= H \Leftrightarrow Hx=Hy;$

therefore, $A_{1}x\neq A_{1}y$ if and only if $Hx\neq Hy$. It follows from this equivalence and the definition of $\mathcal{C}$ that $\mathcal{C}$‘s order $q$ coincides with the number of distinct (right) cosets of $H$ in $G$; i.e., $q=p^{\alpha}r/h$, or $h=p^{\alpha}r/q$. But since $p\nmid q$, this implies that $q\,|\, r$, so that $h\geq p^{\alpha}$

To see the reverse inequality, consider an element $a\in A_1$. The definition of $H$ implies that the (left) coset $aH$ is a subset of $A_1$; whence,

$h=|H|=|aH|\leq |A_1|=p^{\alpha}$,

and it now follows that $h=p^{\alpha}$. Therefore, the subgroup $H$ of $G$ is indeed a member of $\mathcal{S}$

$\Box$

Remark Other authors—e.g., [1, 2]—have presented Wielandt’s proof; Professor Petrich (see below) distilled it particularly nicely on one chalkboard.

## In memoriam

Submitted in honour of Mario Petrich (1932–2021), who showed me (and the rest of the class) this proof at Simon Fraser University in 1986.

## References

[1]  I.N. Herstein, Topics in Algebra, Second edition, Xerox College Publishing, Lexington MA, 1975.

[2]  I. Martin Isaacs, Algebra. A Graduate Course, Brooks/Cole, Pacific Grove CA, 1994.

[3]  Helmut Wielandt, Ein Beweis für die Existenz der Sylowgruppen, Arch. Math. (Basel), 10 (1959), 401–402.

AMS (MOS) Subject Classifications: 20D20, 05A10, 00A05, 11-01

Email the author: mark.kayll@umontana.edu