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.

  • $\begingroup$ 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. $\endgroup$ – Wadim Zudilin Jul 6 '10 at 4:30
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    $\begingroup$ 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 $u(2)$. $\endgroup$ – algori Jul 6 '10 at 4:57
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    $\begingroup$ ..that is $U(n)$ is not a normal subgroup of $SO(2n)$.. $\endgroup$ – algori Jul 6 '10 at 5:03
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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jul 6 '10 at 5:40
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    $\begingroup$ Following algori's observations, shouldn't the title of the question perhaps be changed to remove the word "why"? $\endgroup$ – Yemon Choi Jul 6 '10 at 5:59