How do Subgroup Inverse Maps Work in Group Theory?

[jimmycricket]

Homework Statement

For a group $G$ consider the map $i:G\rightarrow G , i(g)=g^{-1}$
For a subgroup $H\subset G$ show that $i(gH)=Hg^{-1}$ and $i(Hg)=g^{-1}H$

Homework Equations

The Attempt at a Solution

I know that for $g_1,g_2 \in G$ we have $i(g_1g_2)=(g_1g_2)^{-1}=g_2^{-1}g_1^{-1}$
Then since for any $h\in H, h\in G$ we have $i(g_1h)=(g_1h)^{-1}=h^{-1}g_1^{-1}$
Is this a good approach to the problem?

 

[pasmith]

Homework Statement

For a group $G$ consider the map $i:G\rightarrow G , i(g)=g^{-1}$
For a subgroup $H\subset G$ show that $i(gH)=Hg^{-1}$ and $i(Hg)=g^{-1}H$

Homework Equations

The Attempt at a Solution

I know that for $g_1,g_2 \in G$ we have $i(g_1g_2)=(g_1g_2)^{-1}=g_2^{-1}g_1^{-1}$
Then since for any $h\in H, h\in G$ we have $i(g_1h)=(g_1h)^{-1}=h^{-1}g_1^{-1}$
Is this a good approach to the problem?

Working out what $i(gh)$ is for $h \in H$ is certainly a good start.

 

[jimmycricket]

Sorry I should have said I'm actually stuck at this point. Any pointers or hints would be appreciated :)

 

[pasmith]

Sorry I should have said I'm actually stuck at this point. Any pointers or hints would be appreciated :)

You are asked to show that, if $H$ is a subgroup of $G$, then for all $g \in G$, $i(gH) = Hg^{-1}$.

So far you have that if $h \in H$ and $g \in G$ then $i(gh) = h^{-1}g^{-1}$. You now need to explain why $h^{-1}g^{-1} \in Hg^{-1}$.

 

[jimmycricket]

since $H$ is a subgroup, any $h\in H$ has an inverse element $h^{-1}\in H$ such that $hh^{-1}=h^{-1}h=e$ hence $h^{-1}g^{-1}\in Hg^{-1}$