# Subgroup inverse map question

1. Feb 13, 2014

### jimmycricket

1. The problem statement, all variables and given/known data

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 $i(Hg)=g^{-1}H$
2. Relevant equations

3. 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?

2. Feb 13, 2014

### pasmith

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

3. Feb 13, 2014

### jimmycricket

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

4. Feb 13, 2014

### pasmith

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 $h^{-1}g^{-1} \in Hg^{-1}$.

5. Feb 13, 2014

### 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 $h^{-1}g^{-1}\in Hg^{-1}$

Last edited: Feb 13, 2014
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