# Math Help - Subgroups of the orthogonal group O(n)

1. ## Subgroups of the orthogonal group O(n)

I am reading Aigli Papantonopoulou's book Algebra: Pure and applied.

Chapter 14: Symmetries defines the orthogonal group as follows:

O(n, $\mathbb{R}$) = {A $\in$GL(n, $\mathbb{R}$)| $A^T$ = $A^{-1}$} where GL(n, $\mathbb{R}$) is the general linear group.

I need some help with the following problem: Problem 12 of Exercises 14

"Let G be a subgroup of O(n). Show that either every element A $\in$ has det A =1 or exactly half do"

Would appreciate some help with this exercise.

Peter

2. ## Re: Subgroups of the orthogonal group O(n)

Originally Posted by Bernhard
I am reading Aigli Papantonopoulou's book Algebra: Pure and applied.

Chapter 14: Symmetries defines the orthogonal group as follows:

O(n, $\mathbb{R}$) = {A $\in$GL(n, $\mathbb{R}$)| $A^T$ = $A^{-1}$} where GL(n, $\mathbb{R}$) is the general linear group.

I need some help with the following problem: Problem 12 of Exercises 14

"Let G be a subgroup of O(n). Show that either every element A $\in$ has det A =1 or exactly half do"

Would appreciate some help with this exercise.

Peter
Hi Peter!

What can you say about the determinants of $A, A^T, A^{-1}, (A A^{-1})$?

And are you familiar with the sign map, which is a homomorphism to the group {-1, +1}?

3. ## Re: Subgroups of the orthogonal group O(n)

Hi,

Thanks for getting me thinking!

On page 457 Papantonopoulou writes: $\(det A)^2$ = det A det $A^T$ = det( $A^T$A) = det $I_n$ = 1

[as an aside I am not sure why $\(det A)^2$ = det A det $A^T$ as distinct from $\(det A)^2$ = det A det A???]

So det A = 1 or +1

Papantonopoulou also shows that if det a = 1 then A is a rotation of the plane by some angle and if det A = -1 then det A is a reflection.

But where to from here?

Peter

4. ## Re: Subgroups of the orthogonal group O(n)

perhaps you should first prove that $\det A = \det A^T$. here's a hint: in the definition of the determinant, the sgn function is used, so it suffices to show that every term's parity remains unchanged when you "switch the indices" (swaps of matrix terms above the diagonal and below the diagonal occur in pairs...equivalently, since determinants are usually only defined for matrices over commutative rings, "rearranging" the sequence of the multiplicands in each term, amounts to changing:

$\sum_{\sigma \in S_n}\text{sgn}(\sigma)\prod_i^n a_{i\sigma(i)}$ to $\sum_{\sigma \in S_n} \text{sgn}(\sigma^{-1}) \prod_i^n a_{i\sigma^{-1}(i)}$

and the inverse of a permutation has the same disjoint cycle structure (and thus parity) as the original permutation).

now, from the above, and prior posts, we know that the determinant of an orthogonal matrix is either 1 or -1, but you need to show that either HALF or NONE of the matrices in any subgroup of the orthogonal group have determinant -1.

consider the subgroup of O(n,R), SO(n,R). this is a normal subgroup of O(n,R) (check determinants!). since we only have 2 possible determinants for O(n,R), exactly half the elements of O(n,R) have determinant 1 (the other half, which is "the other coset" of SO(n,R), has determinant -1).

now for any subgroup H of O(n,R), H∩SO(n,R) is also a subgroup of O(n,R). show that this is a normal subgroup of H (this should be fairly obvious).
since det:H-->{-1,1} is a homomorphism, what can you say about its kernel?

5. ## Re: Subgroups of the orthogonal group O(n)

Thanks Deveno!!

I am now satisfied that $\det A = det A^T$. Thanks [By the way what text were you taking your notation from?]

You write:

"consider the subgroup of O(n,R), SO(n,R)"

-- How did you know to consider SO(n,R)?

Then you write that SO(n, R) is a normal subgroup and hint that I should check determinants

--- How does checking determinants indicate a normal subgroup?

You then write:

"since we only have 2 possible determinants for O(n,R), exactly half the elements of O(n,R) have determinant 1"

Why does "only two possible determinants" lead to "exactly half the elements of O(n,R) have determinant 1" and finally - what is the link to cosets?

Sorry for all the questions :-(

Peter

6. ## Re: Subgroups of the orthogonal group O(n)

In the previous post I wrote:

"Then you write that SO(n, R) is a normal subgroup and hint that I should check determinants

--- How does checking determinants indicate a normal subgroup?"

================================================== ===

O(n,R) = {A $\in$ GL(n,R) | $A^T = A^{-1}$}

SO(n,R) = {A $\in$ O(n,R) | det A = 1}

Definition

A subgroup H of a group G is normal if $ghg^{-1}$ $\in$ H for all h $\in$ H and g $\in$ G

Take h $\in$ SO(n,R) and g $\in$ O(n,R)

Consider $ghg^{-1}$

det( $ghg^{-1}$) = (det g)(det h)(det $g^{-1}$)

= (det g).1.(1/det g)

= 1

Thus $ghg^{-1}$ $\in$ SO(n,R)

7. ## Re: Subgroups of the orthogonal group O(n)

yes, in fact it is the same way you show SL(n,R) is normal in GL(n,R). the fact that det is a group homomorphism from GL(n,R) to R-{0} turns out to be very useful.

now, about the cosets: we know that det:O(n,R)-->{-1,1} is ONTO. this means that {-1,1} is isomorphic to O(n,R)/SO(n,R), which has 2 cosets, since {-1,1} has two elements.

one of these cosets is simply the kernel SO(n,R), the other is M(SO(n,R)), where M is a matrix with determinant -1 (for example, a reflection consisting of a diagonal matrix with all 1's on the diagonal except for the n,n-th entry, which is -1). as you recall from group theory, all cosets of a quotient group have the same cardinality.

for an arbitrary subgroup H of SO(n,R), we have two options:

H/H∩SO(n,R) is isomorphic to {1}
H/H∩SO(n,R) is isomorphic to {-1,1}.

the first case leads to every element of H has determinant 1, the second leads to exactly half of them have determinant -1.