Let $G$ be a group and $H_{1},H_{2}\leq G $. Then $H_1 \cup H_2 \leq G \iff H_1 \subset H_2$ or $H_2 \subset H_1$ Let $G$ be a group, $H_{1},H_{1} \leq G $. Then $H_1 \cup H_2 \leq G \iff H_1 \subset H_2$ or $H_2 \subset H_1$
I'm stucked at this very trivial proof of groups. Here's my attempt:
$(\Leftarrow)$
$x \in H_{1} \Rightarrow x \in H_{2} $  or $x \in H_{2} \Rightarrow x \in H_{1} $
Let $x,y \in H_1 \cup H_2$
Then $x,y \in H_1$ or $x,y \in H_2$
Then $x.y^{-1} \in H_1 $ or  $x.y^{-1} \in H_2 $ (They're subgroups)
Then $x.y^{-1} \in H_1 \cup H_2$
And $e \in H_1,H_2 \Rightarrow e \in H_1 \cup H_2$
Therefore $H_1 \cup H_2 \subset_{sg} G $
The above I believe is ok, but please verify.
$ (\Rightarrow)$
Let $x \in H_1 ,y \in H_2$
Then $x,y \in H_1 \cup H_2 $
Then $x.y \in H_1 \cup H_2 $ (It's a subgroup)
Then $x.y \in H_1$ or $x.y \in H_2$
Then $y \in H_1$ or $x \in H_2$  ($x^{-1} \in H_1$ and $y^{-1} \in H_2$)
And here is where I'm stucked.
Can someone please help me? Thanks.
 A: For the second part, if $H_{1} \le H_{2}$, it's ok. So suppose $H_{1} \not\le H_{2}$. Then there's $x \in H_{1} \setminus H_{2}$. Let $y \in H_{2}$ be arbitrary. Then $x y \in H_{1} \cup H_{2}$. If $x y \in H_{2}$, then $x = x(x y) y^{-1} \in H_{2}$, against the hypothesis. Then $x y \in H_{1}$, and thus $y = x^{-1} (x y) \in H_{1}$. We have proved that $H_{2} \le H_{1}$.
The first part is easier. If $H_{1} \subseteq H_{2}$, say, then $H_{1} \cup H_{2} = H_{2}$ is a subgroup.

I am using the standard notation $\le$ for what you write $\subset_{sg}$.
A: Only a proof here based on contradiction (see the answer of Andreas for the proof-verification).

Let it be that $H_{1}\cup H_{2}$ is a group.
Assume that $H_{1}\setminus H_{2}$ and $H_{2}\setminus H_{1}$ are
both not empty and let $a\in H_{1}\setminus H_{2}$ and $b\in H_{2}\setminus H_{1}$.
Then $a,b\in H_1\cup H_2$ and consequently $ab\in H_1\cup H_2$.
However
$ab\in H_{2}$ implies that $a=(ab)b^{-1}\in H_{2}$ and $ab\in H_{1}$ implies
that $b=a^{-1}(ab)\in H_{1}$. 
This contradiction shows that it cannot be true that $H_{1}\setminus H_{2}$
and $H_{2}\setminus H_{1}$ are both not empty.
Evidently $H_{1}\setminus H_{2}=\varnothing\iff H_{1}\subseteq H_{2}$
and $H_{2}\setminus H_{1}=\varnothing\iff H_{2}\subseteq H_{1}$.

Conversely if $H_{1}\subseteq H_{2}$ then $H_{1}\cup H_{2}=H_{2}$
hence is a group. 
Likewise if $H_{2}\subseteq H_{1}$ then $H_{1}\cup H_{2}=H_{1}$
hence is a group.
