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 $\displaystyle G:=GL(2,\mathbb{R})\,,\,\,H:=\{A\in G\;;\;\det A=1\}$ ...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
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.
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
(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