# Isomorphism of an orthogonal group.

This question is taken directly from Artin's "Algebra", on page 150:

Is $O_{n}$ isomorphic to the product group of $SO_{n}$ and ${\pm1}$?

Here, $O_{n}$ is defined as the group of orthogonal matrices, $SO_{n}$ is the special group of orthogonal matrices (i.e. orthogonal matrices which have determinant or 1 or -1).

This shouldn't be a hugely difficult problem but for some reason I am confused.

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What property of a product are you having trouble verifying (or disproving)? –  Qiaochu Yuan Oct 1 '12 at 21:47
compare also math.stackexchange.com/questions/29279/… –  Phira Dec 6 '12 at 21:20

When $n$ is odd it is clear we have $O_n \cong SO_n \times \{\pm I\}$ via the product map. Now if $n$ is even, the center of $O_n$ is $\pm I$ while the center of $SO_n$ is also $\pm I$. If $O_n \cong SO_n \times \pm I$ when $n$ is even, they would have isomorphic centers. But then

$$\pm I \not\cong \pm I \times \pm I$$

which is a contradiction. So for $n$ even, $O_n$ is not isomorphic to $SO_n \times \pm I$.

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No. $A\in O_n$ iff the columns of $A$ form an orthonormal system, i.e. $A^TA=I$, see e.g.wiki, hence $\det A=\pm 1$ already there. And $SO_n=\{A\in O_n\mid \det A=1\}$.

Well, if $n$ is even, $\det(-I)=1$, so $A\mapsto (-I)\cdot A$ is invariant in $O_n$..

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For it to be a direct product we need an element of order 2 with determinant -1. Thus the eigenvalues are all $\pm 1$. The diagonal matrix with a single -1 and the rest 1 has the right property. But this does not commute with $SO_n$. In order for it to commute with $SO_n$ we need a scalar matrix. So $O_n$ is a direct product of $SO_n$ and <-1> iff $n$ is odd.

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First consider the case when $n$ is even.

It should be clear that $\\-\mathbb{I}$ and $\mathbb{I} \in O_n$. Since $n$ is even, then $n = 2k | k\in \mathbb{Z}$. Thus $det(-\mathbb{I})=(-1)^{2k}=1$ and $det(\mathbb{I})=1^{2k}=1$. This implies that $\\-\mathbb{I}$ and $\mathbb{I} \in SO_n$. Thus, $SO_n \cap \{\pm\mathbb{I}\}=\{\pm\mathbb{I}\}\neq\mathbb{I}$. This implies that there does not exist an injective function from $SO_n \times \{\pm\mathbb{I}\}$ to $O_n$. Therefore, $O_n$ cannot be isomorphic to $SO_n \times \{\pm\mathbb{I}\}$ for $n$ even.

Now consider the case for $n$ odd.

It should be clear that for $n$ odd, $\\-\mathbb{I} \not\in SO_n$ and $\mathbb{I} \in SO_n$. Thus, $SO_n \cap \{\pm\mathbb{I}\}=\mathbb{I}$. Therefore there exists an injective function from $SO_n \times \{\pm\mathbb{I}\}$ to $O_n$.

We also know that $SO_n$ is a subgroup of $O_n$ of index 2. This means that $SO_n$ is a normal subgroup of $O_n$. It follows that $SO_n \cdot \mathbb{I} =SO_n$ and $SO_n \cdot -\mathbb{I}=O_n-SO_n$ since $SO_n$ has index 2. This implies that $SO_n \cdot \{\pm\mathbb{I}\}=O_n$. It follows that the injective function is a bijection.

Additionally, we know that $\{\pm\mathbb{I}\}\subset Z(GL_n(R))$. Since $\{\pm\mathbb{I}\} \subset O_n$ and $(Z(GL_n(R))\cap O_n) \subset Z(O_n)$, it follows that $\{\pm\mathbb{I}\} \subset Z(O_n)$. This implies that $\{\pm\mathbb{I}\}$ is a normal subgroup of $O_n$. Since $SO_n$ and $\{\pm\mathbb{I}\}$ are normal subgroups of $O_n$, then there exists a homomorphism from $SO_n \times \{\pm\mathbb{I}\}$ to $O_n$.

Therefore, there exists an isomorphism between $O_n$ and $SO_n \times \{\pm\mathbb{I}\}$.

Therefore, $O_n$ is isomorphic to $SO_n \times \{\pm\mathbb{I}\}$ only when $n$ is odd.

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