Question about normal subgroup

Suppose $\displaystyle G$ is a group, and $\displaystyle H$ is some **subset** of $\displaystyle G$.

Is it true that if $\displaystyle \forall g\in G$ and $\displaystyle \forall h\in H$: $\displaystyle ghg^{-1}\in H$, then $\displaystyle H$ is a **normal subgroup**?

Or that $\displaystyle H$ has to be a **subgroup **(not just any subset) for this statment to be true?

I'm asking because I need to prove it, and I not 100 percent sure my teacher didn't make a mistake here...

Thanks in advance.

Re: Question about normal subgroup

Hi Stormey,

Suppose G is abelian and H is __any__ subset of G. Isn't it true that ghg^{-1} is in H for all g in G and all h in H?

Perhaps your teacher wants you to prove that <H>, the subgroup generated by H, is a normal subgroup.

Re: Question about normal subgroup

Quote:

Originally Posted by

**johng** Hi Stormey,

Suppose G is abelian and H is __any__ subset of G. Isn't it true that ghg^{-1} is in H for all g in G and all h in H?

Perhaps your teacher wants you to prove that <H>, the subgroup generated by H, is a normal subgroup.

But it didn't say there that G is abelian, and if G isn't abelian, then this claim isn't true.

So either G is abelian, or H is a subgroup, for this claim to be true, isn't it?

Re: Question about normal subgroup

Stormey,

One counterexample is always enough to refute a statement. However, your statement is also false for non-abelian groups.

Example: Let G=S_{3} be the symmetric group on {1,2,3}, x the 3 cycle (123) and H = {x,x^{-1}}. I leave it to you to prove that ghg^{-1} is in H for all g in G and h in H. "Clearly", H is not a subgroup of G.

Re: Question about normal subgroup

Excellent. :)

This is what I was looking for.

Re: Question about normal subgroup

*edit

well, I found some counter example to this too:

if $\displaystyle H=\left \{(1 \hspace{3} 2 \hspace{3} 3),(3 \hspace{3} 2 \hspace{3} 1) \right \}$, and let's say $\displaystyle g=(1 \hspace{3} 2)$, then:

$\displaystyle ghg^{-1}=(1 \hspace{3} 2)\circ (1 \hspace{3} 2 \hspace{3} 3)\circ (1 \hspace{3} 2)\notin H$

(Worried)

Re: Question about normal subgroup

Quote:

Originally Posted by

**Stormey** *edit

well, I found some counter example to this too:

if $\displaystyle H=\left \{(1 \hspace{3} 2 \hspace{3} 3),(3 \hspace{3} 2 \hspace{3} 1) \right \}$, and let's say $\displaystyle g=(1 \hspace{3} 2)$, then:

$\displaystyle ghg^{-1}=(1 \hspace{3} 2)\circ (1 \hspace{3} 2 \hspace{3} 3)\circ (1 \hspace{3} 2)\notin H$

(Worried)

Actually $\displaystyle (1 ~ 2) \circ (1~2~3) \circ (1~2) = (1~2) \circ (1~3~2) = (1~3~2)$ which is equivalent to (3 2 1).

-Dan

Re: Question about normal subgroup

Oh... I also got (1 3 2), but didn't relize it's equivalent to (3 2 1).

Thanks Dan!

Re: Question about normal subgroup

Quote:

Originally Posted by

**Stormey** Oh... I also got (1 3 2), but didn't relize it's equivalent to (3 2 1).

Thanks Dan!

No problem. Someone had to point that out to me about two weeks ago. (Doh)

-Dan