Hi guys.

I need to prove the following statment, I did most of the work (which I'm not entirely sure is valid - so a feedback would be great), but I'm having some trouble proving the last part of it, and could really use some tips here.

Let $\displaystyle G$ be a group.

Let $\displaystyle H,K$ be non-trivial subgroups of $\displaystyle G$ ($\displaystyle \left \{ e \right \} <H,K<G$).

Prove that $\displaystyle H\cup K\neq G$.

Proof:

If $\displaystyle H=K$ we're done, since $\displaystyle H\cup K=H$, and $\displaystyle H<G$, we get that there exist $\displaystyle g\in G$ s.t $\displaystyle g\notin H\cup K$.

If $\displaystyle H\neq K$, without loss of generality, suppose that $\displaystyle K<H$.

Let $\displaystyle h\neq e\in H$ s.t $\displaystyle h\notin K$ (there exist such $\displaystyle h$ since $\displaystyle H$ isn't trivial and $\displaystyle K<H$).

Let $\displaystyle k\in K$.

Obviusly $\displaystyle h\cdot k\in G$, so If we show that $\displaystyle h\cdot k\notin H\cup K$, we're done.

Let's assume by contradiction that $\displaystyle h\cdot k\in H\cup K$.

It then follows that $\displaystyle h\cdot k\in H \vee h\cdot k\in K$.

Case $\displaystyle I$ ($\displaystyle h\cdot k\in K$):

If $\displaystyle h\cdot k\in K$, then $\displaystyle h\cdot k=m$, for some $\displaystyle m\in K$.

But then:

$\displaystyle h\cdot k=m$

$\displaystyle h\cdot k\cdot k^{-1}=m\cdot k^{-1}$

$\displaystyle h=m\cdot k^{-1}\newline$

but $\displaystyle m\cdot k^{-1}\in K \Rightarrow h\in K$! that contradics our assumption of $\displaystyle h\notin K$.

Case $\displaystyle II$ ($\displaystyle h\cdot k\in H$):

I have nothing here...

any help would be greatly appreciated!