Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Checking if H is a subgroup of G

  1. Jun 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Let [itex]G = GL_n(\mathbb{C})[/itex] and [itex]H = GL_n(\mathbb{R})[/itex]

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

    3. The attempt at a solution

    My intuition says the answer is yes since the [itex]\mathbb{R} \subset \mathbb{C}[/itex], it seems reasonable to say the same about the general linear group of invertible [itex]n \times n[/itex] 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 [itex]a,b \in GL_n(\mathbb{R})[/itex]. Then the product of two [itex]n \times n[/itex] matrices [itex]a,b[/itex] is an [itex]n \times n[/itex] matrix [itex]ab[/itex]. Thus [itex]ab \in GL_n(\mathbb{R})[/itex].

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

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

    So does this show that [itex]GL_n(\mathbb{R}) \subset GL_n(\mathbb{C})[/itex]? I'm confused since I haven't even mentioned [itex]GL_n(\mathbb{C})[/itex].
     
  2. jcsd
  3. Jun 20, 2011 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Yes, you did. You said that [itex]GL_n(\mathbb{R})\subset GL_n(\mathbb{C})[/itex]. Once you have that, you only need to look at [itex]GL_n(\mathbb{R})[/itex].
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook