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 $G$.
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 $A, B$ […]
Nontrivial 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 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 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 $P$ […]