Is the product of all conjugates of some subgroup independent of the order? Let $G$ be a finite group and $A \le G$. Let $A^G = \{ A_1, A_2, \ldots, A_n \}$ be all the conjugates of $A$, i.e. each $A_i$ equals $A^g$ for some $g \in G$. Then I want to show that
$$
 A_1 A_2 \cdots A_n = A_{\pi(1)} A_{\pi(2)} \cdots A_{\pi(n)}
$$
for each permutation $\pi \in \mathcal S_n$. I guess I have to apply $ba = ab^a$, but If I move some element, i.e. consider $a,b \in A$ and $a^g b^h \in A^gA^h$, then $a^g b^h = b^h (a^g)^{b^h}$ and $(a^g)^{b^h}$ need not be in $A^g$? (also in general I guess two conjugates of a subgroup need not commute, i.e. $A^g A^h \ne A^h A^g$ in general, right?).
Any hints (maybe how to apply $ba = ab^a$ correctly...) how to solve it?
 A: You are right in saying that the question you asked is equivalent to the exercise, but I am afraid that I found your formulation, with its emphasis on the conjugates commuting unhelpful, because they need not commute pairwise.
An element of $\langle A_1,\ldots,A_n \rangle$ can be written as a product $x_1x_2\cdots x_l$, where each $x_j$ is in one of the $A_i$. The proof is by induction on $l$. Using $x_jx_k = x_kx_j^{x_k}$, we can move every $x_k$ that lies in $A_1$ to the left in the product. If there were any terms in $A_1$ then, after doing this, the length of the remaining product is less than $l$, so we can put that into the required form by induction, and we are done.
Suppose then that there were no terms in the original product in $A_1$ and that $t$ is minimal such that some term lies in $A_t$. We use induction on $t$, and we have dealt with the case $t=1$. We can move all of the $A_t$ terms to the left in the product and apply induction to the remaining terms to get an expression $a y_1\cdots y_k$, with $a \in A_t$ and $y_1 \cdots y_k$ in the required form. If $y_1 \in A_u$ with $u \ge t$ then we are done.
Otherwise,  $y \in A_u$ with $u<t$, and then the result follows by the induction on $t$. 
