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?

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- Apr 11th 2010, 07:33 AMLightningepsilonIdentity and Inverse of a Permutation set
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? - Apr 11th 2010, 10:54 AMtonio

Uuh?? First you have $\displaystyle G:=GL(2,\mathbb{R})\,,\,\,H:=\{A\in G\;;\;\det A=1\}$ ...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 - Apr 12th 2010, 02:56 AMLightningepsilon
Sorry for the very badly worded question!

What i meant was we have $\displaystyle G:=GL(2,\mathbb{R})\,,\,\,H:=\{A\in G\;;\;\det A=1\}$ 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.

(Worried) - Apr 12th 2010, 03:38 AMtonio
Ok, now I see: well, you have to answer the following questions:

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

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

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

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

Tonio - Apr 12th 2010, 05:26 AMLightningepsilon
(3)

$\displaystyle A,B\in H$ then

$\displaystyle det(AB) = det(A)det(B) = 1 * 1 = 1 $

and so $\displaystyle AB\in H$

(2)

I suppose the inverse must be 'I' and so i have to find $\displaystyle A^{-1}$

where $\displaystyle A*A^{-1} = I$.

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

$\displaystyle 1 = det(I) = det(AA^{-1}) = det(A)det(A^{-1})$

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

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