# Thread: Show GL_2(R) is isomorphic to Z_2 x Z_2

1. ## Show GL_2(R) is isomorphic to Z_2 x Z_2

Let G be the set of matrices $GL_2(R)$ over R.

Show that G is isomorphic to $Z_2 x Z_2$.

I know GL2R represents the matrices

1 0
0 1

-1 0
0 1

1 0
0 -1

-1 0
0 -1

But how can I show it is isomorphic to $Z_2 x Z_2$?

2. Originally Posted by elninio
Let G be the set of matrices $GL_2(R)$ over R.

Show that G is isomorphic to $Z_2 x Z_2$.

I know GL2R represents the matrices

1 0
0 1

-1 0
0 1

1 0
0 -1

-1 0
0 -1

But how can I show it is isomorphic to $Z_2 x Z_2$?

Something doesn't fit in here: if R is the real numbers, then $GL_2(\mathbb{R})$ denotes the group of all invertible 2 x 2 matrices with real entries and this is an infinite group, so it cannot possibly be isomorphic with the Klein group, a group of order 4.

Please do detail what you mean by $GL_2(\mathbb{R})$

Tonio

3. Looks like i've made a mistake in the translation of the question. I misused notation.

The questions direct translation is:

Let G be the following set of matrices over R:

1 0
0 1

-1 0
0 1

1 0
0 -1

-1 0
0 -1

Show that G is isomorphic to Z_2 x Z_2.

This makes more sense, right?

4. There are only two groups (up to isomorphism) of order 4: $\mathbb{Z}_4,$ which is cyclic, and $\mathbb{Z}_2\times\mathbb{Z}_2,$ whose non trivial elements are of order 2.

So showing that the square of any matrix in your group is the identity is sufficient to prove that it is isomorphic to $(\mathbb{Z}_2\times\mathbb{Z}_2,+)$