Let $G$ a non abelian group of order 20 generated by two elements $\sigma$ and $\tau$ such that $G = \langle \sigma, \tau |\,\, \sigma^5 = \tau^4 = 1,\, \tau\sigma = \sigma^3\tau,\, \sigma\tau = \tau\sigma^2 \rangle$.
We have that $G$ acts on a group (H, .) by the following action, $\forall \,a \in H$ and $\forall\, \sigma_1, \sigma_2 \in G$, we have $(a^{\sigma_1})^{\sigma_2} = a^{\sigma_2\sigma_1}$.
Let $H^+ = \{ a \in H \,|\, a^{\tau^2} = a^{\sigma}= a\}$, $H^- = \{ a \in H \,|\, a^{\tau^2} = a^{-1}\}$ and $H^{1-\sigma} = \{ a^{1-\sigma} \,|\, a \in H\}$ be a subgroups of $H$.
I am trying to prove that $(H^+)^{1-\sigma} \subset H^+$ and $(H^-)^{1-\sigma} \subset H^+$ in order to deduce that $H^{1-\sigma} \subset H^+$ because we have that $H = H^+ \times H^-$.
I proced as follows:
Let $a \in H^+$, since $(a^{1-\sigma})^{\tau^2}\,=\,a^{\tau^2-\tau^2\sigma}\,=\,a^{\tau^2-\sigma^4\tau^2}\,=\,(a^{\tau^2})^{1-\sigma^4}\,=\,a^{1-\sigma^4} = a^{1-\sigma}$ because $a^{\sigma^4} = a^{\sigma}$, then $(H^+)^{1-\sigma}\,\subset\, H^+$.
But a referee gave this comments:
"you write $(a^{1-\sigma})^{\tau^2}\,=\,a^{\tau^2-\tau^2\sigma}$. This seems like you are thinking that $(a^{1-\sigma})^{\tau^2}\,=\,a^{\tau^2(1-\sigma)}$. This is not the correct interpretation of the left-hand side. To see the correct interpretation of the left-hand side, notice you are wanting to show that $(H^+)^{1-\sigma} \subset H^+$. Thus you want to show that $\tau^2$ acts trivially on $(H^+)^{1-\sigma}$. You can see what the correct interpretation of the left-hand side has to be from this."
I can't understand where is the mistake in my reasoning, and what is the correct interpretation of the left-hand side.