# Show that $(S, \cdot)$ is a subgroup of $(GL_3(\mathbb{R}), \cdot))$

$(GL_3(\mathbb{R}), \cdot))$ is the group of all invertible real $3 \times 3$ matrices, operation is multiplication. And let $$S= \left\{A \in \mathbb{R}^{3 \times 3} \mid \text{det} A \in \left\{-1,1\right\}\right\}$$

Show that $(S, \cdot)$ is a subgroup of $(GL_3(\mathbb{R}), \cdot))$

Firstly we need to show that $S$ is a subset of $G$. This means we need to show that $S$ has invertible $3 \times 3$ matrix. We know from the set that $S$ is a $3 \times 3$ matrix, also we know that it has only invertible matrices since the determinant is always $\neq 0$. So we already know that $S$ is a subset of $G$.

Secondly we need to show that $S$ is a group.

1. associativity (matrix multiplication always associative, $A(BC) = (AB)C$)

2. neutral element (what to say about that?)

3. inverse element (exists anyway because we know that the set only includes invertible matrices)

So the only thing I'm not sure about is the neutral element. Are the other two, and the rest what I've written fine?

1)the identity matrix has determinant $1$: $\det (I)=1$
2) the determinant of the inverse matrix is the inverse of the determinant: $\det{(A^{-1})}=\left(\det A \right)^{-1}$
• @tenepolis When you prove that some subset $H$ of group $G$ is a group on its own (i.e., subgroup), you must show that for any $g, h \in H$ hold $gh \in H$ and $g^{-1} \in H$. That's what this hint was almost about, how to prove these two properties. – Evgeny Mar 18 '17 at 9:17
• You know that the matrix $A$ has an inverse because its determinant is $1$ or $-1$ (not null). And, since the determinant of the inverse is also $1$ or $-1$, also the inverse matrix is an element of $S$ – Emilio Novati Mar 18 '17 at 9:19