# Commutator subgroup of matrices of order two!

Let $G$ be the group of all invertible matrices on the real field of order two under multiplication. And define $N=\{X\in G:det(X)=1\}$, prove that $N=G'$. That's easy to prove that $G'\subset N$ using simple properties of determinant!
The main part of solution is to prove that $N\subset G'$

Note that $G'$ Denotes the commutator subgroup of $G$.

• When discussing groups of matrices, you need to specify the field. Also, you probably mean degree two rather than order two. – Derek Holt Feb 10 '15 at 20:07
• A solution can be found at many places, e.g., here. – Dietrich Burde Feb 10 '15 at 20:16
• The set of invertible matrices of order 2 is not a group in general. For example, the general linear group has a subgroup isomorphic to the symmetric group, which is generated by elements of order 2 but contains elements that have order different from 2. – Matt Samuel Feb 10 '15 at 20:19
• @MattSamuel I expect the question is referring to $2 \times 2$ matrices, not matrices of order $2$. – Derek Holt Feb 10 '15 at 22:08

Hint: You generate $G'$ by transductions, i.e., shear maps.