Show that if $H$ and $K$ are subgroups of one abelian group $G$, then $G_1=\{hk\mid h\in H\text{ and }k\in K\}$ is a subgroup of $G$ [duplicate]

Here's the problem:

Show that if $H$ and $K$ are subgroups of one abelian group $G$, then $G_1=\{hk\mid h\in H\text{ and } k\in K\}$ is a subgroup of $G$

I know that "abelian" means that you add the commutative property to the definition of the group, but I haven't been able to see how to use it in the problem.

So far, I've tried to suppose that $G_1$ is a subgroup of $G$ and then I tried to prove the three parts of the theorem that is in the book I'm working on: "A subset $H$ of a group $G$ is a subgroup of $G$ if and only if: (i) $H$ is closed un the binary operation of $G$, (ii) the identity $e$ of $G$ is in $H$ and (iii) for every $a\in H$ is true that $a^{-1}\in H$ too.

I got stuck in (i) because I don't know how to prove the closure with the given conditions. I have $h_0k_0, h_1k_1 \in G_1 $ but how do I prove that $h_0k_0 * h_1k_1 \in G_1 $? And how can I continue to the identity?

Any help would be appreciated. Thanks for reading.

4 Answers

Already there is a good answer above, so I try to provide a general, famous theorem in abstract algebra. The theorem statement is,

If $H$ and $K$ are subgroups of a group $G$ (not necessarily abelian) such that $K$ is a normal subgroup of $G$, then the set $HK$ is a subgroup of $G$.

Proof: Pick two elements $h_{1}k_{1}$ and $h_{2}k_{2}$ of $HK$. Then, $$h_{1}k_{1}(h_{2}k_{2})^{-1}=h_{1}k_{1}k_{2}^{-1}h_{2}^{-1}=(h_{1}h_{2}^{-1})\cdot(h_{2}k_{1}h_{2}^{-1})\cdot(h_{2}k_{2}^{-1}h_{2}^{-1}) \in HK$$ since $h_{1}h_{2}^{-1} \in H$ and $(h_{2}k_{1}h_{2}^{-1})\cdot(h_{2}k_{2}^{-1}h_{2}^{-1}) \in K$ by the definition of the normal subgroup.

If you accept this theorem, then your question is obviously solved since any subgroup of an abelian group is normal.

I hope my answer is helpful for you to get a wider viewpoint.

We gonna prove the point $i)$ of a subgroup chracterization.

$i)$ Let $x,y\in G_1$ then $x=h_1k_1$ and $y=h_2k_2$ where $h_1h_2\in H$ and $k_1,k_2\in K$ then $$xy=(h_1k_1)(h_2k_2)=(h_1h_2)(k_1k_2)$$ since $G$ is connmutative and $H,K$ are subgroups of $G$ then as a subset their inherit the property of be connmutative.

hence $G_1$ is closed under the sum.

Now for the identity, let us denote by $1_G$ be the identity of $G$, since $H,K$ are subgroups of $G$ their have the identity of $G$ it is $1_G\in H$ and $1_G\in K$ now what product of $h_1k_1$ you should consider to get $1_G$?

Easy since we can use $1_G1_G=1_G\in G_1$ .

for the least propery $iii)$ (the existence of inverses) you can read the Shaun Post.

Use the one-step subgroup test. The standard notation for $G_1$ is $HK$.

Since $e\in H$ and $e\in K$, we have $e=ee\in HK$. Thus $HK\neq\varnothing$.

Let $g=hk\in HK$ for $h\in H$ and $k\in K$. Then, since $H\le G$, $K\le G$, and $G$ is closed under multiplication, we have $g\in G$. Hence $HK\subseteq G$.

Let $f=ab, g=hk\in HK$ for $a,h\in H$, $b,k\in K$. Then

$$\begin{align} fg^{-1}&=(ab)(hk)^{-1}\\ &=abk^{-1}h^{-1}\\ &=(ah^{-1})(bk^{-1}), \end{align}$$

but $ah^{-1}\in H$ as $H\le G$ and $bk^{-1}\in K$ as $K\le G$. Hence $fg^{-1}\in HK$.

Hence $HK\le G$.

Closure under addition:

$$(h_1k_1)(h_2k_2)=h_1(k_1h_2)k_2 = h_1(h_2k_1)k_2=(h_1h_2)(k_1k_2)\in HK $$ Closure under inverse: $$ (hk)^{-1} = k^{-1}h^{-1}=h^{-1}k^{-1}\in HK$$ Thus HK is a subgroup.