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Let $G$ and $H$ be two groups and let $$\alpha : H~\times~G \rightarrow H$$ be a right action of $G$ on $H$ so that $$\alpha_{g}(h) = h\prime$$ where $g\in G$, $h, h\prime\in H.$ I want to write the inverse of $h\prime$ in terms of $g$ and $h$. Please I want references on where I can get this information.

Dziedzorm
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  • The way you write it looks like it should be a RIGHT action of $G$ on $H$ (you wrote left). Also, does $G$ act by automorphisms of $H$? – Nick Nov 27 '20 at 03:17
  • Hi @Nick, G does not act as an automorphism of H – Dziedzorm Nov 27 '20 at 06:39
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    @Dziedzorm I don't know that there is much that you can say about this without knowing the action. If $G$ acting by right multiplication, $\alpha_{g}(h) = hg = h^{\prime}$ and so $(h^{\prime})^{-1} = g^{-1}h^{-1}$. But even then you can't write it in terms of the group action without knowing some relation involving $h$ and $g$. – xxxxxxxxx Nov 27 '20 at 06:50
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    Aren't group actions required to be homomorphisms? So why isn't $$ h^{\prime-1}=\alpha_g\big(h^{-1}\big)\ ? $$ – lonza leggiera Nov 27 '20 at 06:56
  • @lonzaleggiera Not in the way you are suggesting. In fact $g_1 g_2$ acts in the same way as first $g_1$ then $g_2$. – ancient mathematician Nov 27 '20 at 07:34
  • @ancientmathematician I don't understand. If $\ \alpha_g\ $ is a homomorphism, isn't it true that $$ Id_H=\alpha_g\big(hh^{-1}\big)=\alpha_g(h)\alpha_g\big(h^{-1}\big)=h^\prime\alpha_g\big(h^{-1}\big)\ ? $$ If so, then multiplying this equation on the left by $\ h^{\prime-1}\ $ gives $\ h^{\prime-1}=\alpha_g\big(h^{-1}\big)\ $. – lonza leggiera Nov 27 '20 at 07:44
  • $\alpha_g$ is not a homomorphism, what is true is that $\alpha_{g_1 g_2}=\alpha_{g_1}\alpha_{g_2}$; that's what we know about a group action. – ancient mathematician Nov 27 '20 at 09:08
  • Thank you for the explanation. I didn't read Wikipedia's definition of group action carefully enough. – lonza leggiera Nov 27 '20 at 20:55
  • @MorganRodgers The equation $(hg)^{-1}=h^{-1}g^{-1}$ assumes writing $g$ and $h$ next to each other refers to a group operation, but it doesn't here, it refers to a group action. – anon Nov 28 '20 at 04:05
  • In general, OP, if the action of $G$ on $H$'s underlying set has nothing to do with $H$'s group operation, then the inverse of $hg$ (wrt $H$'s group operation) cannot be expressed using $G$'s action on $H$ and the elements $h$ and $g$. Is there more context you're not telling us? – anon Nov 28 '20 at 04:06
  • @runway44 You say "writing $g$ and $h$ next to each other... doesn't [refer to group multiplication] here" - this notation is not used at all by the OP, I introduced it in my comment; so yes, in fact it does refer to group multiplication here. – xxxxxxxxx Nov 28 '20 at 06:12
  • @runway44 This is a common way to have a group acting on itself, particularly for actions on cosets. And it does in fact describe a (right) group action, right actions are a little strange since we want $\alpha_{h}(\alpha_{g}(x)) = \alpha_{gh}(x)$ for these (if we wanted a left action we would need to define $\alpha_{g}(x) = xg^{-1}$). – xxxxxxxxx Nov 28 '20 at 06:36

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