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I came to this problem when doing the exercise that the polydisc $\Delta(0,1)^n=\prod\limits_{n}\Delta(0,1)$ in $\mathbb{C}^n$ is not biholomorphic to the unit ball $\mathbb{B}^n$ in $\mathbb{C}^n$.We can show that the group of automorphisms of $\Delta(0,1)^n$ preserving the origin is the semidirect of $U^n(1)=\prod\limits_{n}U(1)$ and the permutation group $S_n$(the semidirect product here means that $U^n(1)$ si a normal subgroup of $U^n(1)\rtimes S_n$), while that of $\mathbb{B}^n$ is $U(n)$.

Then I do not know how to show that the above two groups are not isomorphic.Intuitively, topologically the (real) dimension of $U^n(1)\rtimes S_n$ is $n$ , while that of $U(n)$ is $\frac{n(n-1)}{2}$ (is it right?), so they are not isomrphic.

Clearly it is not a rigorous proof.

Will someone be kind enough to give me some hints for this problem?Thank you very much!

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One simple enough way is to count the square roots of the identity. The $n=2$ case is illustrative. The equation $((z_1,z_2),\sigma)^2 = ((z_1^2,z_2^2),\sigma^2) = ((1,1),e)$ gives us 8 square roots of the identity in $U^2(1)\rtimes S_2$, while it is easily verified that there are only 6 square roots of the identity matrix in $U(2)$. Generalizing to all $n>1$ shouldn't be too hard.

Alternatively, you could set up a map $U^n(1)\rightarrow U(n)$ and look at where tuples with all but one entry equal to $1$ are sent, then verify that this cannot be a normal subgroup. This may require more creativity (I haven't worked it out), but would also be prettier.

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:thank you very much!I got it. – user14242 Dec 24 '11 at 9:14

You can use the natural $n \times n$ matrix representation of $U(n).$ Sketch: This is clearly irreducible. By Clifford's theorem, every Abelian normal subgroup is still completely reducible, so diagonalizable via a unitary matrix. Hence we may suppose that $U(1)^n$ is embedded as a normal subgroup consisting of diagonal matrices. In fact, it must then be the group of all diagonal matrices in $U(n).$ The normalizer in $U(n)$ of the group of diagonal matrices permutes the weight spaces (simultaneous eigenspaces of all diagonal matrices), and there are $n$ of these. Hence the normalizer of $U(1)^n$ consists of monomial matrices (that is, matrices with one non-zero entry in each row and column). When $n > 1,$ there are clearly unitary matrices which are not monomial.

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