Is U(n) a normal subgroup of SO(2n)?

U(n) is the group of n by n unitary complex matrices and SO(2n) is the group of 2n by 2n real orthogonal matrices with determinant 1.So far I can show that how to get an injective group homomorphism from U(n) to SO(2n). This shows that U(n) is isomorphic to a subgroup of SO(2n).But I have no idea whether U(n) is normal in SO(2n)? If someone could explain this to me or just point out some reference for me,I will appreciate your help. Thanks.

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Have you checked with the references at en.wikipedia.org/wiki/Classical_group ? By the way, your tagging is misleading: group-theory and lie-groups are definitely more efficient. –  Wadim Zudilin Jul 6 '10 at 4:30
One of the reasons this is so difficult may be that $U(n)$ is not a normal subgroup of $SO(n)$ at all (unless $n=1$): when $n>2$ the Lie algebra $so(2n)$ is simple and does not have any ideals and when $n=2$ the only ideals are copies of $su(2)$, not $u(2)$. –  algori Jul 6 '10 at 4:57
..that is $U(n)$ is not a normal subgroup of $SO(2n)$.. –  algori Jul 6 '10 at 5:03
Finding an injection from a group $A$ to a group $B$ doesn't show that $A$ is a subgroup of $B$; it shows that $A$ is isomorphic to a subgroup of $B$. The distinction is important (in general, though I don't know about the particular case you ask about), e.g., the group $D_8$ of symmetries of a square has several subgroups isomorphic to the cyclic group of order 2, but one of those subgroups is normal in $D_8$ and the others aren't. –  Gerry Myerson Jul 6 '10 at 5:40
Following algori's observations, shouldn't the title of the question perhaps be changed to remove the word "why"? –  Yemon Choi Jul 6 '10 at 5:59

Thanks. The answers clarify many things for me. The reason that I ask this question is that I am confused about what's the meaning of the quotient $SO(2n)/U(n)$ if $U(n)$ is not isomorphic to a normal subgroup of $SO(2n)$?
The quotient SO(2n)/U(n) simply means to take the (right) coset space, which is NOT a group in this case. However, using the natural map $\pi:SO(2n)\rightarrow SO(2n)/U(n)$, we put a topology on $SO(2n)/U(n)$ which turns $SO(2n)/U(n)$ into a (topological) manifold. Further, we can declare $\pi$ to be a submersion which puts a differentiable structure on $SO(2n)/U(n)$, making it a smooth manifold. (All of this applies equally well to any $G/H$ with $G$ and $H$ compact. If things are not compact, one must be a little careful.) –  Jason DeVito Jul 6 '10 at 12:53
Huichi -- if $G$ is a group and $H$ is a subgroup of $G$, normal or not, one can consider the set $G/H$ formed by the left cosets $gH,g\in G$. If $G$ is a Lie group and $H$ is a closed subgroup (hence, automatically a Lie subgroup), then $G/H$ is a smooth manifold of dimension $\dim G-\dim H$. –  algori Jul 6 '10 at 13:04
@Jason: I know that's equivalent, but isn't $G/H$ technically the left coset space $\{aH\}$? The left cosets are the ones that have a left action of $G$. –  Greg Kuperberg Jul 6 '10 at 17:46