Example of Two Groups and a Subgroup of the Direct Product that is Not of the Form of Direct Product

Problem 467

Give an example of two groups $G$ and $H$ and a subgroup $K$ of the direct product $G\times H$ such that $K$ cannot be written as $K=G_1\times H_1$, where $G_1$ and $H_1$ are subgroups of $G$ and $H$, respectively.

Let $G$ be any nontrivial group, and let $G=H$.
(For example, you may take $G=H=\Zmod{2}$.)

Then consider the subset $K$ in the direct product given by
\[K:=\{(g,g) \mid g\in G\} \subset G\times G.\]

We claim that $K$ is a subgroup of $G\times G$.
In fact, we have
\begin{align*}
(g,g)(h,h)=(gh,gh)\in K \text{ and }\\
(g,g)^{-1}=(g^{-1}, g^{-1})\in K
\end{align*}
for any $g, h\in G$.
Thus, $K$ is closed under multiplications and inverses, and hence $K$ is a subgroup of $G\times G$.

Now we show that $K$ is not of the form $G_1\times H_1$ for some subgroups $G_1, H_1$ of $G$.
Assume on the contrary $K=G_1\times H_1$ for some subgroups $G_1, H_1$ of $G$.

Since $G$ is a nontrivial group, there is a nonidentity element $x\in G$.
So $(x,x)\in K$ and $K$ is not the trivial group.
Thus, both $G_1$ and $H_1$ cannot be the trivial group.

Without loss of generality, assume that $G_1$ is nontrivial.
Then $G_1$ contains a nonidentity element $y$.

Since the identity element $e$ is contained in all subgroups, we have
\[(y,e)\in G_1\times H_1.\]
However, this element cannot be in $K$ since $y\neq e$, a contradiction.

Abelian Group and Direct Product of Its Subgroups
Let $G$ be a finite abelian group of order $mn$, where $m$ and $n$ are relatively prime positive integers.
Then show that there exists unique subgroups $G_1$ of order $m$ and $G_2$ of order $n$ such that $G\cong G_1 \times G_2$.
Hint.
Consider […]

Non-Abelian Group of Order $pq$ and its Sylow Subgroups
Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.
Hint.
Use Sylow's theorem. To review Sylow's theorem, check […]

Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups
Let $G$ be an abelian group and let $f: G\to \Z$ be a surjective group homomorphism.
Prove that we have an isomorphism of groups:
\[G \cong \ker(f)\times \Z.\]
Proof.
Since $f:G\to \Z$ is surjective, there exists an element $a\in G$ such […]

Eckmann–Hilton Argument: Group Operation is a Group Homomorphism
Let $G$ be a group with the identity element $e$ and suppose that we have a group homomorphism $\phi$ from the direct product $G \times G$ to $G$ satisfying
\[\phi(e, g)=g \text{ and } \phi(g, e)=g, \tag{*}\]
for any $g\in G$.
Let $\mu: G\times G \to G$ be a map defined […]

Isomorphism Criterion of Semidirect Product of Groups
Let $A$, $B$ be groups. Let $\phi:B \to \Aut(A)$ be a group homomorphism.
The semidirect product $A \rtimes_{\phi} B$ with respect to $\phi$ is a group whose underlying set is $A \times B$ with group operation
\[(a_1, b_1)\cdot (a_2, b_2)=(a_1\phi(b_1)(a_2), b_1b_2),\]
where $a_i […]

A Condition that a Commutator Group is a Normal Subgroup
Let $H$ be a normal subgroup of a group $G$.
Then show that $N:=[H, G]$ is a subgroup of $H$ and $N \triangleleft G$.
Here $[H, G]$ is a subgroup of $G$ generated by commutators $[h,k]:=hkh^{-1}k^{-1}$.
In particular, the commutator subgroup $[G, G]$ is a normal subgroup of […]

A Simple Abelian Group if and only if the Order is a Prime Number
Let $G$ be a group. (Do not assume that $G$ is a finite group.)
Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.
Definition.
A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is […]

If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup
Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.
Hint.
It follows from […]