Troubling calculation for a wreath product of groups.

Let $G$ be a group, $H$ a transformation group acting on a set $S$, and suppose $G$ acts on another set $T$. Let $G\wr H$ denote the wreath product of $G$ and $H$. So the composition for $(f_1,h_1),(f_2,h_2)\in G\wr H$, (where $f$ is a map of $S\to G$), is defined as $$(f_1,h_1)(f_2,h_2)=(f_1(h_1f_2),h_1h_2).$$ Also, if $G^S$ denotes the set of maps, it is a group if we define $(f_1f_2)(s)=f_1(s)f_2(s)$. And for $h\in H$ and $f\in G^S$, define $hf$ by $(hf)(s)=f(h^{-1}s)$, which is an action of $H$ on $G^S$. I wanted to check that for $(t,s)\in T\times S$, the rule $(f,h)(t,s)=(f(s)t,hs)$ gives an action of $G\wr H$ on $T\times S$.

I calculate \begin{align*} [(f_1,h_1)(f_2,h_2)](t,s) &= (f_1(h_1f_2),h_1h_2)(t,s)\\ &= ((f_1(h_1f_2))(s)t,(h_1h_2)(s))\\ &= ((f_1(s)(h_1f_2)(s))t,h_1h_2s). \end{align*}

But \begin{align*} (f_1,h_1)[(f_2,h_2)(t,s)] &= (f_1,h_1)(f_2(s)t,h_2s)\\ &= (f_1(h_2(s))f_2(s)t,h_1h_2s)\\ \end{align*} which is puzzling since I'm getting different $h_i$ in the first entry, when they should be equal, to satisfy one of the properties of being a group action. Have I applied something incorrectly here?

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I think, maybe you should regard the side you take the functions. I mean that sometimes we write $xf$ and sometimes we take $f(x)$. Actions may lead to different results. – Babak S. Jun 2 '12 at 6:44
Right, but do you see an error in my calculations? – Hailie Mathieson Jun 2 '12 at 7:00
I write your action like $(t,s)^{(f,h)}$=$(t^{f(s)},s^h)$ wherein $(t,s)\in T× S$ so, $((t,s)^{(f_1,h_1)})^{(f_2,h_2)}$=$((t^{f_1(s)},s^{h_1}))^{(f_2,h_2)}$=$(t^{f_1(‌​s)f_2(s^{h_1}),s^{h_1h_2}})$. I think this notation makes clear what you are looking for. Hope it help. – Babak S. Jun 2 '12 at 8:00
The wreath product symbol is typset with \wr, rather than with \int. – Arturo Magidin Jun 2 '12 at 22:12

This is exercise 1.12.11 out of Jacobsons Basic Algebra 1 right? I believe that this question has a mistake in the phrasing. Given you notation, one should define the action of $G\wr H$ on $T\times S$ by $$(f, h)(t, s):= (f(hs)t, hs).$$ Doing this, you avoid the problems you were experiencing in the calculation.
Use \wr for the wreath product symbol. \int produces the wrong symbol (mirror image and the shading is all wrong), and bad spacing, – Arturo Magidin Jul 8 '12 at 2:02