# $(\mathbb C^{\times}, \cdot)$ is a subgroup of $(GL(n,\mathbb C), \cdot)$

I am learning groups and subgroups in my algebra course. Today, we talked about examples of subgroup but I am not sure why the following holds:

$(\mathbb C^{\times}, \cdot)$ is a subgroup of $(GL(n,\mathbb C), \cdot)$.

For clarity, I am considering the multiplicative group of the set of complex numbers(excluding $0$) and I want to show that it is a subgroup of $(GL(n,\mathbb C), \cdot)$, the set of all invertible $n\times n$ matrices with complex entries, together with matrix multiplication as the group operation.

I thought about diagonal matrix, but I am in a mess in writing the formal proof of the statement. Thanks in advance.

You are correct that you can look at diagonal matrices. Simply consider the set $D = \{ aI | a \in \mathbb{C}^\ast\}$ where $I$ is the identity matrix. Then you see:
• $I \in D$, thus the identity is contained in $D$.
• $aI, bI \in D$ then we have that $aI \cdot bI = abI \in D$ thus $D$ is closed under matrix multiplication.
• If $aI \in D$ then $a^{-1}I \in D$ hence the inverse is in $D$.
• Thank you for replying. Why the above proof can be used to conclude that $(\mathbb C_{\neq 0}, \cdot)$ is a subgroup? Didn't we just prove D is a subgroup? Commented Sep 19, 2014 at 8:28
• It is worth noting that sometimes mathematicians are sloppy about what "is" a subgroup. In this case, in the usual sense, $\mathbb{C}^\times$ is not strictly a subgroup. However, by the above argument it is isomorphic to a subgroup (and in a pretty natural way), so we often abuse notation/terminology and just say that it is a subgroup. Commented Sep 19, 2014 at 20:16