# The Product Distinct Sylow $p$-Subgroups Can Never be a Subgroup

## Problem 544

Let $G$ a finite group and let $H$ and $K$ be two distinct Sylow $p$-group, where $p$ is a prime number dividing the order $|G|$ of $G$.

Prove that the product $HK$ can never be a subgroup of the group $G$.

## Hint.

Use the following fact.

If $H$ and $K$ are finite subgroups of a group $G$, then we have
$|HK|=\frac{|H| |K|}{|H \cap K|}.$

## Proof.

Let $p^{\alpha}$ is the highest power of $p$ that divides $|G|$.
That is, we have
$|G|=p^{\alpha}n,$ where $p$ does not divide the integer $n$.

Then the orders of the Sylow $p$-subgroups $H, K$ are $p^{\alpha}$.

Since the intersection $H\cap K$ is a subgroup of $H$, the order of $H \cap K$ is $p^{\beta}$ for some integer $\beta \leq \alpha$ by Lagrange’s theorem.
As $H$ and $K$ are distinct subgroups, we must have $\beta < \alpha$.

Then the number of elements of the product $HK$ is
\begin{align*}
|HK|&=\frac{|H| |K|}{|H \cap K|}\6pt] &=\frac{p^{\alpha} p^{\alpha}}{p^{\beta}}=p^{2\alpha-\beta}. \end{align*} Since \beta < \alpha, we have 2\alpha-\beta > \alpha. It follows that the product HK cannot be a subgroup of G since otherwise the order |HK|=p^{2\alpha-\beta} divides |G| by Lagrange’s theorem but p^{\alpha} is the highest power of p that divides G. Sponsored Links ### More from my site • Fundamental Theorem of Finitely Generated Abelian Groups and its application In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem. Problem. Let G be a finite abelian group of order n. If n is the product of distinct prime numbers, then prove that G is isomorphic […] • Use Lagrange’s Theorem to Prove Fermat’s Little Theorem Use Lagrange's Theorem in the multiplicative group (\Zmod{p})^{\times} to prove Fermat's Little Theorem: if p is a prime number then a^p \equiv a \pmod p for all a \in \Z. Before the proof, let us recall Lagrange's Theorem. Lagrange's Theorem If G is a […] • Group of Order pq is Either Abelian or the Center is Trivial Let G be a group of order |G|=pq, where p and q are (not necessarily distinct) prime numbers. Then show that G is either abelian group or the center Z(G)=1. Hint. Use the result of the problem "If the Quotient by the Center is Cyclic, then the Group is […] • Group of Order pq Has a Normal Sylow Subgroup and Solvable Let p, q be prime numbers such that p>q. If a group G has order pq, then show the followings. (a) The group G has a normal Sylow p-subgroup. (b) The group G is solvable. Definition/Hint For (a), apply Sylow's theorem. To review Sylow's theorem, […] • If Two Subsets A, B of a Finite Group G are Large Enough, then G=AB Let G be a finite group and let A, B be subsets of G satisfying \[|A|+|B| > |G|. Here $|X|$ denotes the cardinality (the number of elements) of the set $X$. Then prove that $G=AB$, where $AB=\{ab \mid a\in A, b\in B\}.$   Proof. Since $A, B$ […]
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