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.

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 $G$.

The post The Product of Distinct Sylow $p$-Subgroups Can Never be a Subgroup first appeared on Problems in Mathematics.

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 $A, B$ are subsets of the group $G$, we have $AB\subset G$.
Thus, it remains to show that $G\subset AB$, that is any element $g\in G$ is of the form $ab$ for some $a\in A$ and $b\in B$.
This is equivalent to finding $a\in A$ and $b\in B$ such that $gb^{-1}=a$.

Consider the subset
\[B^{-1}:=\{b^{-1} \mid b \in B\}.\] Since taking the inverse gives the bijective map $B \to B^{-1}$, $b \mapsto b^{-1}$, we have $|B|=|B^{-1}|$.

Also consider the subset
\[gB^{-1}=\{gb^{-1} \mid b\in B\}.\] Note that multiplying by $g$ and by its inverse $g^{-1}$ give the bijective maps
\[B^{-1} \to gB^{-1}, b^{-1} \mapsto gb^{-1} \text{ and } gB^{-1} \to B^{-1}, gb^{-1} \mapsto b^{-1}.\] Hence we have
\[ |B|=|B^{-1}|=|gB^{-1}|.\]

Since $A$ and $gB^{-1}$ are both subsets in $G$ and we have by assumption that
\[|A|+|gB^{-1}|=|A|+|B| > |G|,\] the intersection $A\cap gB^{-1}$ cannot be empty.

Therefore, there exists $a \in A\cap gB^{-1}$, and thus $a\in A$ and $a=gb^{-1}$ for some $b\in B$.
As a result we obtain $g=ab$.
It yields that $G\subset AB$, and we have $G=AB$ as a consequence.

Related Question.

As an application, or use the similar technique, try the following

See the post↴
Each Element in a Finite Field is the Sum of Two Squares
for a proof of this problem.

The post If Two Subsets $A, B$ of a Finite Group $G$ are Large Enough, then $G=AB$ first appeared on Problems in Mathematics.

Problem 448

Let $G$ be a group. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$.
The product of $H$ and $N$ is defined to be the subset
\[H\cdot N=\{hn\in G\mid h \in H, n\in N\}.\] Prove that the product $H\cdot N$ is a subgroup of $G$.

Definition.

A subgroup $N$ of a group $G$ is called a normal subgroup if for any $g\in G$ and $n\in N$, we have
\[gng^{-1}\in N.\]  

Proof.

We prove that the product $H\cdot N$ is closed under products and inverses.

Let $h_1n_1$ and $h_2n_2$ be elements in $H\cdot N$, where $h_1, h_2\in H$ and $n_1, n_2\in N$.
Let $e$ be the identity element in $G$.
We have
\begin{align*}
(h_1n_1)(h_2n_2)&=h_1en_1h_2n_2\\
&=h_1(h_2h_2^{-1})n_1h_2n_2 && \text{since $h_2h_2^{-1}=e$}\\
&=(h_1h_2)(h_2^{-1}n_1h_2n_2). \tag{*}
\end{align*}
Since $H$ is a subgroup, the element $h_1h_2$ is in $H$.
Also, since $N$ is a normal subgroup, we have $h_2^{-1}n_1h_2$ is in $N$. Hence
\[h_2^{-1}n_1h_2n_2=(h_2^{-1}n_1h_2)n_2\in N.\] It follows from (*) that the product
\[(h_1n_1)(h_2n_2)=(h_1h_2)(h_2^{-1}n_1h_2n_2)\in H\cdot N.\] Therefore, the product $H\cdot N$ is closed under products.

Next, let $hn$ be any element in $H\cdot N$, where $h\in H$ and $n\in N$.
Then we have
\begin{align*}
(hn)^{-1}&=n^{-1}h^{-1}\\
&=en^{-1}h^{-1}\\
&=(h^{-1}h)n^{-1}h^{-1} &&\text{since $h^{-1}h=e$}\\
&=h^{-1}(hn^{-1}h^{-1}).
\end{align*}
Since $N$ is a normal subgroup, we have $hn^{-1}h^{-1}\in N$, and hence
\[(hn)^{-1}=h^{-1}(hn^{-1}h^{-1})\in H\cdot N.\] Thus, the product $H\cdot N$ is closed under inverses.

This completes the proof that the product $H\cdot N$ is a subgroup of $G$.

The post The Product of a Subgroup and a Normal Subgroup is a Subgroup first appeared on Problems in Mathematics.

Let $G, G’$ be groups and let $f:G \to G’$ be a group homomorphism.
Put $N=\ker(f)$. Then show that we have
\[f^{-1}(f(H))=HN.\]

Proof.

$(\subset)$ Take an arbitrary element $g\in f^{-1}(f(H))$. Then we have $f(g)\in f(H)$.
It follows that there exists $h\in H$ such that $f(g)=f(h)$.
Since $f$ is a group homomorphism, we obtain
\[f(h^{-1}g)=e’,\] where $e’$ is the identity element of the group $G’$.

This implies that $h^{-1}g\in \ker(f)=N$, hence we have
\begin{align*}
g\in hN\subset HN.
\end{align*}
Therefore we have $f^{-1}(f(H)) \subset HN$.

$(\supset)$ On the other hand, let $g\in HN$ be an arbitrary element.
Then we can write $g=hn$ with $h \in H$ and $n\in N$.
We have
\begin{align*}
f(g)&=f(hn)=f(h)f(n)\\
&=f(h)e’=f(h)\in f(H)
\end{align*}
since $f$ is a group homomorphism and $f(n)=e’$.
Thus we have
\[g\in f^{-1}(f(H))\] and $f^{-1}(f(H)) \supset HN$.

Therefore, putting the two continents together gives
\[f^{-1}(f(H))=HN\] as required.

The post Group Homomorphism, Preimage, and Product of Groups first appeared on Problems in Mathematics.