# Thread: HH = H, etc.

1. ## HH = H, etc.

"H is a subgroup of G, a is in H. Show that Ha = H."

Is this as simple as showing that since H is a group, all operations are closed within that group? Or is there more to it than that? If it's the latter, I'm stuck.

2. ## Re: HH = H, etc.

Originally Posted by phys251
"H is a subgroup of G, a is in H. Show that Ha = H."
Is it clear to you that $\displaystyle Ha=\{x*a:~x\in H\}~?$ (of course $\displaystyle *$ is the operation in $\displaystyle G$)

Is it clear that $\displaystyle Ha\subseteq H~?$ Why?

Can you show $\displaystyle H\subseteq Ha~?$

3. ## Re: HH = H, etc.

Originally Posted by Plato
Is it clear to you that $\displaystyle Ha=\{x*a:~x\in H\}~?$ (of course $\displaystyle *$ is the operation in $\displaystyle G$)
Right, that's just the definition of Ha.

Is it clear that $\displaystyle Ha\subseteq H~?$ Why?
Because of closure?

Can you show $\displaystyle H\subseteq Ha~?$
So the trick is to show that the sets contain each other. Is $\displaystyle H\subseteq Ha$ again a direct result of closure, or is there more to it than that?

4. ## Re: HH = H, etc.

Originally Posted by phys251
So the trick is to show that the sets contain each other. Is $\displaystyle H\subseteq Ha$ again a direct result of closure, or is there more to it than that?
$\displaystyle \\\text{If }h\in H \text{ is it true that }h*a^{-1}\in H~?\text{ WHY?}\\\text{So is it true that }(h*a^{-1})*a\in Ha~?$

5. ## Re: HH = H, etc.

Originally Posted by Plato
$\displaystyle \\\text{If }h\in H \text{ is it true that }h*a^{-1}\in H~?\text{ WHY?}\\\text{So is it true that }(h*a^{-1})*a\in Ha~?$
Wait, this may be simpler than I realize. Since h*a always maps to an element in H, via the properties of groups, HH always maps to H.