Identity and Inverse of a Permutation set

**Lightningepsilon** · April 11th 2010, 07:33 AM

G = GL(2,R), H = {A 2 G : det(A) = 1}

(where the set of all permutations which send 1 → 1)

How could i possibly determine the identity (and consequently the inverse) of the above?

**tonio** · April 11th 2010, 10:54 AM

Originally Posted by Lightningepsilon

G = GL(2,R), H = {A 2 G : det(A) = 1}

(where the set of all permutations which send 1 → 1)

How could i possibly determine the identity (and consequently the inverse) of the above?

Uuh?? First you have $G:=GL(2,\mathbb{R})\,,\,\,H:=\{A\in G\;;\;\det A=1\}$ ...a group of matrices and a subgroup of this group.

And then you write something about permutations which send 1 to 1, and you ask about the identity and inverse "of the above": what "above"?...What is going on??

Tonio

**Lightningepsilon** · April 12th 2010, 02:56 AM

Originally Posted by tonio

Uuh?? First you have $G:=GL(2,\mathbb{R})\,,\,\,H:=\{A\in G\;;\;\det A=1\}$ ...a group of matrices and a subgroup of this group.

And then you write something about permutations which send 1 to 1, and you ask about the identity and inverse "of the above": what "above"?...What is going on??

Tonio

Sorry for the very badly worded question!

What i meant was we have $G:=GL(2,\mathbb{R})\,,\,\,H:=\{A\in G\;;\;\det A=1\}$ and i have to determine whether or not H is a subgroup of G.

I've determined H is closed - now i have to find out whether or not H has an identity and an inverse. Neither or which i can find.

**tonio** · April 12th 2010, 03:38 AM

Originally Posted by Lightningepsilon

Sorry for the very badly worded question!

What i meant was we have $G:=GL(2,\mathbb{R})\,,\,\,H:=\{A\in G\;;\;\det A=1\}$ and i have to determine whether or not H is a subgroup of G.

I've determined H is closed - now i have to find out whether or not H has an identity and an inverse. Neither or which i can find.

Ok, now I see: well, you have to answer the following questions:

1) Is H non-empty? (almost trivial)

2) If $A\in H$ then $A^{-1}\in H$ ?

3) If $A,B\in H$ then $AB\in H$ ?

Hint for (2)-(3): $\det(AB)=\det A\cdot \det B$

Tonio

**Lightningepsilon** · April 12th 2010, 05:26 AM

Originally Posted by tonio

Ok, now I see: well, you have to answer the following questions:

1) Is H non-empty? (almost trivial)

2) If $A\in H$ then $A^{-1}\in H$ ?

3) If $A,B\in H$ then $AB\in H$ ?

Hint for (2)-(3): $\det(AB)=\det A\cdot \det B$

Tonio

(3)
$A,B\in H$ then
$det(AB) = det(A)det(B) = 1 * 1 = 1$
and so $AB\in H$

(2)
I suppose the inverse must be 'I' and so i have to find $A^{-1}$
where $A*A^{-1} = I$ .

So as $A\in H$ then $det(A) = 1$

$1 = det(I) = det(AA^{-1}) = det(A)det(A^{-1})$
hence $det(A^{-1}) = 1$ so $A^{-1}\in H$ .

I can't thank you enough Tonio for helping me, I really appreciate it

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