Abelianization of subgroups of $GL(n,\mathbb{C})$ and $SL(n,\mathbb{C})$ Consider the following subgroups of $GL(n,\mathbb{C})$ and $SL(n,\mathbb{C})$
$$H_G = \{A\in GL(n,\mathbb{C})\: | \: AA^t = \lambda_AI, \text{for some } \lambda_A\in\mathbb{C}^{*}\}\subset GL(n,\mathbb{C})$$
$$H_S = \{A\in SL(n,\mathbb{C})\: | \: AA^t = \lambda_AI, \text{for some } \lambda_A\in\mathbb{C}^{*}\}\subset SL(n,\mathbb{C})$$
Note that $H_S = H_G\cap SL(n,\mathbb{C})$ and that for $H_S$ the complex numbers $\lambda_A$ are $n$-th roots of unity.
So we have a morphism $H_S\rightarrow \mathbb{Z}/n\mathbb{Z}$ mapping $A\mapsto \lambda_A$. The abelianization of $H_S$ is $\mathbb{Z}/n\mathbb{Z}$.
On the other hand, the determinant gives a morphism $H_G\rightarrow \mathbb{C}^{*}$. So on this side it seems that the abelianization of $H_G$ is $\mathbb{C}^{*}$.
Note that $GL(n,\mathbb{C})/H_G\cong SL(n,\mathbb{C})/H_S$. In particualr, the character groups of $H_G$ and $H_S$ should coincide. But this is not the case for what I said before. One of them is $\mathbb{Z}$ and the other one $\mathbb{Z}/n\mathbb{Z}$.
I wanted to ask you if someone can see what I am doing wrong here.
Thanks a lot.
 A: There are two homomorphisms $H_G\to\mathbf{C}^*$, namely $\delta=\det|_{H_G}$ and $\lambda$, and $H_S=\operatorname{Ker}\delta$. First consider $f=\delta\times\lambda: H_G\to\mathbf{C}^*\times\mathbf{C}^*$.
By definition, the kernel of $f$ (which is also the kernel of $\lambda|_{H_S}$) is $\mathrm{SO}(n,\mathbf{C})$. This is known to be a perfect group for every $n$. It follows that $f:H_G\to f(H_G)$ and $\lambda:H_S\to\lambda(H_S)$ are the abelianization homomorphism.
It remains to determine these images. For $n=0$ all groups are trivial, so assume $n\ge 1$.
Then in restriction to $\mathbf{C}^*I_n$, both $\delta$ and $\lambda$ are surjective, given by $\delta(tI_n)=t^n$ and $\lambda(tI_n)=t^2$. Hence, if $(u,v)$ belongs to $f(H_G)$, and $v=w^2$, then $(uw^{-n},1)$ belongs to the image as well, i.e., is the determinant of some orthogonal matrix, which forces $uw^{-n}=\pm 1$, i.e., $u^2=v^n$.
Conversely, if $u^2=v^n$, consider the scalar matrix $wI_n$. Then $f(wI_n)=(w^n,w^2)$. If $w^n=u$, then this equals $(u,v)$ and we are done. Otherwise, $w^n=-u$. Then pick a matrix $M$ in $\mathrm{O}(n)\smallsetminus\mathrm{SO}(n)$: then $f(wM)=(u,v)$.
Hence the image of $f$ is $\{(u,v)\in\mathbf{C}^*\times\mathbf{C}^*:u^2=v^n\}$. If $n=2m$ is even, this is the same as $\{(u,v)\in\mathbf{C}^*\times\mathbf{C}^*:u=\pm v^m\}$ isomorphic to $\mathbf{C}^*\times C_2$ ($C_2$ cyclic of order $2$ generated by $(-1,1)$ and the identity component being $\{u=v^m\}$. If $n=2m+1$ is odd, this abelianization is isomorphic to $\mathbf{C}^*$, parameterized as $\{(w^n,w^2):w\in\mathbf{C}^*\}$ (where $w=uv^{-m}$).
If we restrict to determinant 1, the condition $u^2=v^n$ becomes $u=1$, $v^n=1$. So the image of the abelianization $\lambda:H_S\to\mathbf{C}^*$ is just the group $\mu_n$ of $n$-roots of unity.
