• November 28th 2013, 10:31 AM
Stormey
Suppose $G$ is a group, and $H$ is some subset of $G$.
Is it true that if $\forall g\in G$ and $\forall h\in H$: $ghg^{-1}\in H$, then $H$ is a normal subgroup?
Or that $H$ 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...

• November 28th 2013, 03:57 PM
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.
• November 28th 2013, 11:45 PM
Stormey
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?
• November 29th 2013, 05:16 AM
johng
Stormey,

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

Example: Let G=S3 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.
• November 30th 2013, 12:18 AM
Stormey
Excellent. :)
This is what I was looking for.
• November 30th 2013, 12:50 AM
Stormey
*edit

well, I found some counter example to this too:

if $H=\left \{(1 \hspace{3} 2 \hspace{3} 3),(3 \hspace{3} 2 \hspace{3} 1) \right \}$, and let's say $g=(1 \hspace{3} 2)$, then:
$ghg^{-1}=(1 \hspace{3} 2)\circ (1 \hspace{3} 2 \hspace{3} 3)\circ (1 \hspace{3} 2)\notin H$
(Worried)
• November 30th 2013, 04:30 AM
topsquark
Quote:

Originally Posted by Stormey
*edit

well, I found some counter example to this too:

if $H=\left \{(1 \hspace{3} 2 \hspace{3} 3),(3 \hspace{3} 2 \hspace{3} 1) \right \}$, and let's say $g=(1 \hspace{3} 2)$, then:
$ghg^{-1}=(1 \hspace{3} 2)\circ (1 \hspace{3} 2 \hspace{3} 3)\circ (1 \hspace{3} 2)\notin H$
(Worried)

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

-Dan
• November 30th 2013, 05:17 AM
Stormey
Oh... I also got (1 3 2), but didn't relize it's equivalent to (3 2 1).
Thanks Dan!
• November 30th 2013, 05:44 AM
topsquark