Prove that SL_n(F) is a subgroup of GL_n(F)

The Mathematician — Fri, 09 Dec 2022 13:00:31 +0000

Let

F

be any field. Prove that

SL_n(F)

is a subgroup of

GL_n(F)

.

Proof. Let

F

be any field. Recall that the general linear group is defined as:

\begin{align*} GL_n(F) = \{A \ | \ A \text{ is an } n \times n \text{ invertible matrix with entries from } F\} \end{align*}

and the special linear group is defined as:

\begin{align*} SL_n(F) = \{A \in GL_n(F) \ | \ det(A) = 1\}. \end{align*}

We want to show that if

A,B \in SL_n(F)

, then

A^{-1} \in SL_n(F)

and

AB \in SL_n(F)

. Assume for the first case that

A \in SL_n(F)

. Since

det(A) = 1

, and therefore inverse, we have that:

\begin{align*} det(A^{-1}) = \frac{1}{det(A)} = \frac{1}{1} = 1. \end{align*}

A^{-1} \in SL_n(F)

. For the second case, assume that

A,B \in SL_n(F)

. Then

det(A) = det(B) = 1

. Now we have that:

\begin{align*} det(AB) = det(A)det(B) = 1. \end{align*}

Therefore,

AB \in SL_n(F)

.

So

SL_n(F)

is a subgroup of

GL_n(F)

SL_n(F) is a subgroup of GL_n(F) Archives - Epsilonify