I'm not usually one to post unworked problems here... I usually try to at least have an attempt, but unfortunately in this case I'm unable to even get an intuitive sense of what's going on here - and me being primarily a physicist, I'm pretty paralyzed without some intuition of what's going on. Additionally, this question was set in the homework with zero class notes (i.e. we were not even told what a wreath product is, but now we're proving it's properties)... so we're definitely a little stumped. Any help would be appreciated!

The question:

Let $K$ and $L$ be groups, and let $M$ be the set of functions from $L$ to $K$. We define an action of $L$ on $M$ by setting, for $l \epsilon L$ and $f \epsilon M$:

$^{l}f:\, L\rightarrow K;\,\, x\mapsto f(l^{-1}x)$

We regard M as a group by pointwise multiplication, that is, $(f\cdot f')(x)=f(x)f'(x)$ for all $f, f' \epsilon M$. The elements of the (unrestricted) wreath product $K \wr L$ are pairs $(f,l)$, where $f \epsilon M$ and $l \epsilon L$. Given $(f_{1},l_{1}),(f_{2},l_{2})\,\epsilon\, K \wr L$, we define:


Prove that $K \wr L$ is a group. (Amongst other things, you will need to show that $^{1_{1}}(^{l_{2}}f_{3})=^{(l_{1}l_{2})}f_{3}$ for $l_{1},l_{2}\epsilon L$ and $f_{3}\epsilon M$)

That last part in brackets is a hint given on the sheet by the lecturer. There are two more parts after this question, but if I can't do this, I won't be able to do the others! I am currently working through to try and prove the group axioms - Closure, identity, inverses and associativity, however it is very slow going with little understanding (but I'm trying!)

Thanks all :)



1) Let's prove that $^{l_1} (^{l_2}f) = ^{l_1 l_2}\!\!f$. So, for every $x \in L$ we have $$ ^{l_1} (^{l_2}f)(x)=^{l_2}\!\!f(l_1^{-1}x)=f(l_2^{-1}l_1^{-1}x)=f((l_1 l_2)^{-1}x)=^{l_1 l_2}\!\!f(x). $$

2) Now you should show that for every $(f_{1},l_{1}),(f_{2},l_{2}),(f_{3},l_{3})\,\in\, K \wr L$ the following equality holds: $$ [(f_{1},l_{1})\cdot(f_{2},l_{2})]\cdot(f_{3},l_{3})=(f_{1},l_{1})\cdot[(f_{2},l_{2})\cdot(f_{3},l_{3})] $$ (the associativity of the operation $\cdot$ in $K \wr L$ that is defined as $(f_{1},l_{1})\cdot(f_{2},l_{2})\,:=(f_{1}\cdot^{l_{1}}\!\!f_{2},l_{1}l_{2})$).

3) What is the identity in $K \wr L$?

4) What pair is the inverse of a pair $(f,l) \in K \wr L$?

  • $\begingroup$ That's awesome, thanks! I was able to prove almost all of this, except inverses. For the given operation, it's easy enough to show that the second element of $(f,l)$ has an inverse, but I'm struggling to find an inverse for that first $f$ with the group action on it. $\endgroup$ – Yoshi May 7 '14 at 0:01
  • $\begingroup$ @Yoshi: Try $((^{l^{-1}}\!f)^{-1},l^{-1})$ for inverse, here $f^{-1}$ is the function s.t. $f^{-1}(x)=(f(x))^{-1}$ for every $x \in L$. $\endgroup$ – user35603 May 7 '14 at 6:31
  • $\begingroup$ Thanks! I actually managed to work this out this morning in a burst of inspiration, but its comforting to know you had the same solution! $\endgroup$ – Yoshi May 7 '14 at 7:18


Let $L$ be the circle $S^1$ and let $K$ be a line $\mathbb{R}^1$. Think of the cylinder $L \times K$: a function from $L$ to $K$ has a graph, and it is like a path on the cylinder. Think of individual elements of $L$ as keeping track of the rotation. Then $K \wr L$ consists of pairs $(f,l)$. Given another pair $(f',l')$ we can multiply first by $f'$ and then by $l'$. Multiplying by $f'$ has to pass through $l$, and the result ${}^{l} f'$ is $f'$ rotated by $l$. This is then added to $f$ pointwise. To multiply by $l'$ we just combine the two rotations (so 45 degrees with 45 degrees is 90 degrees).

If you replace "all functions" with "all smooth functions" then you get a subgroup that is easier to visualize.

An example element: $(\sin(x), 45^\circ)$ -- view it as the graph of $\sin(x)$ on the cylinder and a torque looking arrow labelled $45^\circ$.

An example multiplication: $(\sin(x), 45^\circ) (\sin(2x), 15^\circ) = (\sin(x) + \sin(2(x-45^\circ)), 60^\circ)$.


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