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

Group Theory Problems and Solutions

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 $G$.

 
LoadingAdd to solve later

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 $G$.


LoadingAdd to solve later

More from my site

  • Group of Order $pq$ is Either Abelian or the Center is TrivialGroup 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 TheoremUse 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 SolvableGroup 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 57Determine 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 $G=AB$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 $A, B$ […]
  • Nontrivial Action of a Simple Group on a Finite SetNontrivial Action of a Simple Group on a Finite Set Let $G$ be a simple group and let $X$ be a finite set. Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$. Then show that $G$ is a finite group and the order of $G$ divides $|X|!$. Proof. Since $G$ acts on $X$, it […]
  • Normal Subgroup Whose Order is Relatively Prime to Its IndexNormal Subgroup Whose Order is Relatively Prime to Its Index Let $G$ be a finite group and let $N$ be a normal subgroup of $G$. Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$. (a) Prove that $N=\{a\in G \mid a^n=e\}$. (b) Prove that $N=\{b^m \mid b\in G\}$.   Proof. Note that as $n$ and […]
  • Sylow Subgroups of a Group of Order 33 is Normal SubgroupsSylow Subgroups of a Group of Order 33 is Normal Subgroups Prove that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.   Hint. We use Sylow's theorem. Review the basic terminologies and Sylow's theorem. Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Group Theory
Group Theory Problems and Solutions
The Normalizer of a Proper Subgroup of a Nilpotent Group is Strictly Bigger

Let $G$ be a nilpotent group and let $H$ be a proper subgroup of $G$. Then prove that $H \subsetneq...

Close