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}