# The Product of 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. ### More from my site • 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 […] • 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 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, […] • Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57 Let G be a group of order 57. Assume that G is not a cyclic group. Then determine the number of elements in G of order 3. Proof. Observe the prime factorization 57=3\cdot 19. Let n_{19} be the number of Sylow 19-subgroups of G. By […] • 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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