# Thread: Need help with a proof - subgroups

1. ## Need help with a proof - subgroups

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 $G$ be a group.
Let $H,K$ be non-trivial subgroups of $G$ ( $\left \{ e \right \} ).
Prove that $H\cup K\neq G$.

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

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

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

Let $k\in K$.

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

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

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

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

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

But then:

$h\cdot k=m$

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

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

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

Case $II$ ( $h\cdot k\in H$):
I have nothing here...

any help would be greatly appreciated!

2. ## Re: Need help with a proof - subgroups

Originally Posted by Stormey
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 $G$ be a group.
Let $H,K$ be non-trivial subgroups of $G$ ( $\left \{ e \right \} ).
Prove that $H\cup K\neq G$.

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

If $H\neq K$, without loss of generality, suppose that $K.
I see reason that K would have to be a subgroup of H. Consider the symmetries of a square, D_8. H = {e, reflection over the x - axis} and K = {e, reflection over the y - axis} are both subgroups of D_8, but K is not a subgroup of H.

-Dan

3. ## Re: Need help with a proof - subgroups

Originally Posted by topsquark
I see reason that K would have to be a subgroup of H. Consider the symmetries of a square, D_8. H = {e, reflection over the x - axis} and K = {e, reflection over the y - axis} are both subgroups of D_8, but K is not a subgroup of H.

-Dan
Hi, topsquark, thanks for the help.
Yes of course, K does not have to be a subgroup of H.
but there has to be an element in H that doesn't appear in K (or vice versa).
so besides this distinction, the rest of the proof is valid.
(correct me if I'm wrong)

4. ## Re: Need help with a proof - subgroups

[QUOTE=Stormey;804944]
Let $G$ be a group.
Let $H,K$ be non-trivial subgroups of $G$ ( $\left \{ e \right \} ).
//

5. ## Re: Need help with a proof - subgroups

Originally Posted by Stormey
Let $G$ be a group.
Let $H,K$ be non-trivial subgroups of $G$ ( $\left \{ e \right \} ).
Prove that $H\cup K\neq G$.
I am confused by the notation.
I would assume that ( $\left \{ e \right \} means that each of $K~\&~H$ are proper subgroups.

If that is the case, then it must be true that neither is a subset of the other.
So $\left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]$

Now can you argue that $hk \in G$ BUT $hk \notin H \wedge hk \notin K$.

Did I miss-understand that notation?

6. ## Re: Need help with a proof - subgroups

Originally Posted by Plato
I am confused by the notation.
I would assume that ( $\left \{ e \right \} means that each of $K~\&~H$ are proper subgroups.

If that is the case, then it must be true that neither is a subset of the other.
So $\left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]$

Now can you argue that $hk \in G$ BUT $hk \notin H \wedge hk \notin K$.

Did I miss-understand that notation?
No, you got it. that is indeed the meaning of this notation.

Why can I argue that $\left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]$? how do I justify it?

7. ## Re: Need help with a proof - subgroups

Originally Posted by Stormey
No, you got it. that is indeed the meaning of this notation.
Why can I argue that $\left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]$? how do I justify it?
That is the very meaning of $K \not\subset H \wedge H \not\subset K$, surely you know that.

If $K \subseteq H \Rightarrow \;K \cup H = H \ne G$ so neither is the subset of the other.

If $hk \in H \Rightarrow \;hk = h' \Rightarrow \;k = {h^{ - 1}}h' \in H$.

Same for $hk \in K \Rightarrow \;h = k'{k^{ - 1}} \in K$

8. ## Re: Need help with a proof - subgroups

ohh...

So I actually have three cases:

I: where $H=K$.
II: where $H\subset K\vee K\subset K$.
III where $K \not\subset H \wedge H \not\subset K$.

Thank you, again.