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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. –  B. S. Jun 2 '12 at 6:44
    
Right, but do you see an error in my calculations? –  hmIII 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. –  B. 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
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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.

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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
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