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 $\displaystyle G$ 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!

Re: Need help with a proof - subgroups

Quote:

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 $\displaystyle G$ 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 $\displaystyle K<H$.

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

Re: Need help with a proof - subgroups

Quote:

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)

Re: Need help with a proof - subgroups

[QUOTE=Stormey;804944]

Let $\displaystyle G$ be a group.

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

//

Re: Need help with a proof - subgroups

Quote:

Originally Posted by

**Stormey** 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 $\displaystyle H\cup K\neq G$.

I am confused by the notation.

I would assume that ($\displaystyle \left \{ e \right \} <H,K<G$ means that each of $\displaystyle K~\&~H$ are **proper** subgroups.

If that is the case, then it must be true that neither is a subset of the other.

So $\displaystyle \left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]$

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

Did I miss-understand that notation?

Re: Need help with a proof - subgroups

Quote:

Originally Posted by

**Plato** I am confused by the notation.

I would assume that ($\displaystyle \left \{ e \right \} <H,K<G$ means that each of $\displaystyle K~\&~H$ are **proper** subgroups.

If that is the case, then it must be true that neither is a subset of the other.

So $\displaystyle \left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]$

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

Did I miss-understand that notation?

No, you got it. that is indeed the meaning of this notation.

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

Re: Need help with a proof - subgroups

Quote:

Originally Posted by

**Stormey** No, you got it. that is indeed the meaning of this notation.

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

That is the very meaning of $\displaystyle K \not\subset H \wedge H \not\subset K$, surely you know that.

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

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

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

Re: Need help with a proof - subgroups

ohh...

So I actually have three cases:

I: where $\displaystyle H=K$.

II: where $\displaystyle H\subset K\vee K\subset K$.

III where $\displaystyle K \not\subset H \wedge H \not\subset K$.

Thank you, again. (Happy)