Checking if H is a subgroup of G

[Samuelb88]

Homework Statement

Let $G = GL_n(\mathbb{C})$ and $H = GL_n(\mathbb{R})$

Homework Equations

Prop. A subset H of a group G is a subgroup if:
1. If $a,b \in H$, then $ab \in H$.
2. $1 \in H$
3. $\forall a \in H$ $\exists a^{-1} \in H$.

The Attempt at a Solution

My intuition says the answer is yes since the $\mathbb{R} \subset \mathbb{C}$, it seems reasonable to say the same about the general linear group of invertible $n \times n$ matrices. So if I am understanding how to check if H is a subgroup of G correctly, I need to test the three conditions above. Here goes:

1. Let $a,b \in GL_n(\mathbb{R})$. Then the product of two $n \times n$ matrices $a,b$ is an $n \times n$ matrix $ab$. Thus $ab \in GL_n(\mathbb{R})$.

2. The $n \times n$ identity matrix $I \in GL_n(\mathbb{R})$ since $\det(I)=1$ $\forall n$.

3. From the definition of $GL_n(\mathbb{R}) = \{ n \times n \, \, \, invertible \, \, \, real \, \, \, matrices\}$, it follows that $\forall a \in GL_n(\mathbb{R})$ $\exists a^{-1} \in GL_n(\mathbb{R})$.

So does this show that $GL_n(\mathbb{R}) \subset GL_n(\mathbb{C})$? I'm confused since I haven't even mentioned $GL_n(\mathbb{C})$.

 

[HallsofIvy]

Yes, you did. You said that $GL_n(\mathbb{R})\subset GL_n(\mathbb{C})$. Once you have that, you only need to look at $GL_n(\mathbb{R})$.