$|AxA|$ is a constant implies normal subgroup I have an algebra problem:
Let $G$ be a finite group, and $A \le G$. Show that if $|AxA|$ is a constant for all $x \in G$, then $A$ is a normal subgroup of $G$.
My first question is about the meaning of the notation $AxA$. Is this:
$$ \{a_1xa_2 | a_1, a_2 \in A\}?$$
Then what would the proof strategy be?
 A: This is a very interesting problem.  Note that $A$ itself is a double coset, since $A = A1_GA$.  The basic answer is that $A$ is the smallest possible double coset.
For $x \in G$, define a map $\phi: A \rightarrow AxA$ by $\phi(a) = ax$.  This is clearly injective, and if $xAx^{-1} \neq A$, you can show that $\phi$ is not surjective.  This implies that if $A$ is not a normal subgroup of $G$, then there is at least one double coset with more elements than $A$, which gives you what you want.
Extra stuff I find interesting: I came up with this approach by thinking about a beautiful result from linear algebraic groups called the Bruhat decomposition.  Let $F$ be a field, and let $G = \textrm{GL}_n(F)$, the group of $n$ by $n$ matrices with entries in $F$ whose determinant is not zero.  Let $B$ be the group of upper triangular matrices with determinant not zero, and for each permutation $\sigma \in S_n$, let $w_{\sigma}$ be the permutation matrix corresponding to $\sigma$.  For example, if $n = 2$, and $\sigma = (1 \, 2)$, then
$$w_{\sigma} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$
The Bruhat decomposition theorem says that $G$ is equal to the disjoint union of $n!$ double cosets
$$\bigcup\limits_{\sigma \in S_n} Bw_{\sigma}B$$
Since you're dealing with finite groups, let's say $F$ is a finite field, say $F = \mathbb{Z}/p\mathbb{Z}$, then $G$ is a finite group, and these double cosets have different sizes. You can explicitly compute the cardinality of each double coset as follows (though it would take awhile to explain why this works):
Let $i, j$ be unequal integers between $1$ and $n$.  Call a pair $(i,j)$ positive if $i < j$, and negative if $i > j$.  For $\sigma \in S_n$, define the length $\ell(\sigma)$ to be the number of negative elements in the set
$$\{(\sigma(1), \sigma(2)), (\sigma(2), \sigma(3)), ... , (\sigma(n-1),\sigma(n)) \}$$
Then the number of elements in the double coset $Bw_{\sigma}B$ will be $$|Bw_{\sigma}B| = (p-1)^n p^{\frac{(n-1)n}{2} + \ell(\sigma)}$$
There is only one element of length $0$, which is the identity permutation, corresponding to the double coset $B$, and there is only one element $\tau$ of length $n$, which is the permutation $(1 \, n)(2 \, (n-1)) \cdots$, i.e.
$$w_{\tau} = \begin{pmatrix} & & 1 \\ & \dots & \\ 1 & & \end{pmatrix}$$
so the smallest double coset is $B$, the largest double coset is $Bw_{\tau}B$, and all the other double cosets have sizes in between.
Even without knowing the formula for computing the number of elements in a double coset $Bw_{\sigma}B$, you can still tell that the double cosets $Bw_{\sigma}B$ for $\sigma \neq 1$ have more elements than $B$ as follows: it is a general fact that $B$ coincides with its normalizer, that is, if $g \in G$, but not in $B$, then $gBg^{-1} \neq B$.  None of the permutation matrices $w_{\sigma}$ are in $B$ except $w_1 = I_n$.  So the argument I gave above tells you that $B$ will have fewer elements than $Bw_{\sigma}B$ when $\sigma \neq 1$.
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is Problem 41 on p.49 in Herstein's book.
I solved this problem as follows:

$\#AxA=o(AeA)=o(AA)=o(A)$ must hold by the assumption of this problem.
By Problem 40 on p.49 in Herstein's book, $$\#AxB=\frac{o(A)o(B)}{o(A\cap xBx^{-1})}$$ for all subgroups $A, B$ of $G$ and for all elements $x$ of $G$.
We use this formula.
Then, $$o(A)=\#AxA=\frac{o(A)o(A)}{o(A\cap xAx^{-1})}$$ holds for all elements $x$ of $G$.
So, $o(A)=o(A\cap xAx^{-1})$ must hold for all elements $x$ of $G$.
Since $A\cap xAx^{-1}\subset A$, $A\cap xAx^{-1}=A$ must hold for all elements
$x$ of $G$.
This implies $A\subset xAx^{-1}$ for all elements $x$ of $G$.
Let $g$ be an arbitrary element of $G$.
Then, $A\subset gAg^{-1}$.
Let $y\in gAg^{-1}$.
Then, we can write $y=gag^{-1}$ for some $a\in A$.
We can write $a=g^{-1}a^{'}g$ for some $a^{'}\in A$ since $A\subset g^{-1}Ag$.
Then, $y=gag^{-1}=gg^{-1}a^{'}gg^{-1}=a^{'}\in A$.
So, $gAg^{-1}\subset A$.
So, $A=gAg^{-1}$ holds for all elements $g$ of $G$.

