# Math Help - Proving a subgroup relation

1. ## Proving a subgroup relation

Heres the question:
Let G be a group a element in G and H is a subgroup of G. Define K={aha(inverse) such that h element in H. Prove K is a subgroup of G.

Heres what i got.
Let x,y elements in K then there exists c,d elements in H such that x=aca(inverse) and y = ada(inverse). then xy equals (aca(inverse))(ada(inverse)) group the a and a inverse in the middle to get a(cd)a(inverse) c,d elements in H so thats in K.

This is where i get stuck. We know the identity is in H because H subgroup of G how do we show e is in K. Also how do we find the inverse of this and prove its in K.

2. Hi,

$e \in H$ so $aea^{-1} \in K$ (according to the definition of $K$) but $aea^{-1}=e$.

$b \in K$ means that there exists $c\in H$ such that $b=aca^{-1}$. We have $c^{-1} \in H$ so $ac^{-1}a^{-1}\in K$. But $ac^{-1}a^{-1} = (aca^{-1})^{-1}= b^{-1}$.

3. Originally Posted by ChrisBickle
Heres the question:
Let G be a group a element in G and H is a subgroup of G. Define K={aha(inverse) such that h element in H. Prove K is a subgroup of G.

Heres what i got.
Let x,y elements in K then there exists c,d elements in H such that x=aca(inverse) and y = ada(inverse). then xy equals (aca(inverse))(ada(inverse)) group the a and a inverse in the middle to get a(cd)a(inverse) c,d elements in H so thats in K.

This is where i get stuck. We know the identity is in H because H subgroup of G how do we show e is in K. Also how do we find the inverse of this and prove its in K.
Identity: $aea^{-1} = aa^{-1} = e$
Inverse: Consider $b =ah^{-1}a^{-1}$. Can you see why $b\in K$?

4. It is still alittle fuzzy to me...lookin at the post above i see what he is doing and i understand the conclusion but i dont get why (ac)inverse(a)inverse = (aca(inverse))inverse...is it a property of something im missing?

5. Originally Posted by ChrisBickle
It is still alittle fuzzy to me...lookin at the post above i see what he is doing and i understand the conclusion but i dont get why (ac)inverse(a)inverse = (aca(inverse))inverse...is it a property of something im missing?
Either by using the fact that $(ab)^{-1} = b^{-1}a^{-1} \Rightarrow (aca^{-1})^{-1} = ((ac)(a^{-1}))^{-1} = (a^{-1})^{-1}(ac)^{-1} = ac^{-1}a^{-1}$

Or simply going by the fact that $(aca^{-1})(ac^{-1}a^{-1}) = (ac^{-1}a^{-1})(aca^{-1}) = e$