Let $su(N)$ denote the Lie algebra of $SU(N)$, $gl(N,R)$ that of $GL(N,R)$ and so on.
Explain why
$gl(N,R) \sim u(1) \oplus sl(N,R)$,
$ gl(N,C) \sim u(1) \oplus u(1) \oplus sl(N,C)$
$u(N) \sim u(1) \oplus su(N)$, where $\sim$ is isomorphic.
Attempt:
Please criticise any inprecision in the following: The generators of $SL(N,R)$ are all independent $N \times N$ real unit determinant matrices. Same goes for $GL(N,R)$ without the restriction to unit determinant. The generators of $U(N)$ are $\left\{1,T_a\right\}$ where $T_a$ are generators of $SU(N)$, so $U(N)$ now includes the the $N \times N$ unit determinant matrix. $U(1)$ is the only group with lie algebra with the unit determinant matrix as the only generator. I'm not sure what the next step is to infer the isomorphism from these statements. It works on dimensions but why is it also a direct sum of the groups and not say a direct product for example?
For the second example, why are there two copies of $u(1)$?
Many thanks!
You may just work on the level of Lie algebras, which is much easier than the group level because we only deal with vector spaces. The Lie algebra $\mathfrak{sl}(n)$ has dimension $n^2-1$, so $$ \mathfrak{gl}(n)=\mathbb{R}\oplus \mathfrak{sl}(n) $$ as real vector spaces. However, both $\mathfrak{sl}(n)$ and $\mathbb{R}=Z(\mathfrak{gl}(n))=\mathfrak{u}(1)$ are also Lie algebra ideals, hence the first statement follows. In the second example we need two copies of the real algebra $\mathfrak{u}(1)$ because the center of $\mathfrak{gl}_n(\mathbb{C})$ equals $\mathbb{C}=\mathbb{R}\oplus\mathbb{R}=\mathfrak{u}(1)\oplus \mathfrak{u}(1)$.
Edit: The center of a Lie algebra $L$ is given by $Z(L)=\{x\in L\mid [x,L]=0\}$. For example, the center of $\mathfrak{gl}_n(K)$ is $1$-dimensional, generated by the identity. It is a Lie algebra ideal.