3
$\begingroup$

Is the set $GL_n(\mathbb{R})$ is cyclic group?

It is not, right? $GL_1(\mathbb{R})$ is just $\mathbb{R}^{*}$ which is dense. I forget, but there was maybe a result which said something like any cyclic subgroup of Real number must be discrete? Please help.

3 Answers 3

20

Any cyclic group is finite or countable. So $GL_n(\Bbb R)$ is only cyclic if it is trivial, which happens for $n=0$ only.

15

Because a cyclic group has a single generator, it is abelian. For $n\geq2$, $GL_n(\mathbb R)$ is not abelian: $ \begin{bmatrix}1&0\\0&-1 \end{bmatrix}\,\begin{bmatrix}1&1\\1&2 \end{bmatrix}=\begin{bmatrix} 1&1\\-1&-2\end{bmatrix}\ne\begin{bmatrix} 1&-1\\1&-2\end{bmatrix}=\,\begin{bmatrix}1&1\\1&2 \end{bmatrix}\,\begin{bmatrix}1&0\\0&-1 \end{bmatrix} $ As $GL_1(\mathbb R)=\mathbb R$ is uncountable, it cannot be cyclic either.

  • 0
    Your fix is unsavoury. Firstly, it is precisely what Marc can Leeuwen said in his answer. Secondly, it works for all $n$ so makes the rest of your answer redundant. Thirdly, such a statement should be given with proof! I like your answer, but I think you should leave out the case of $n=1$.2012-11-22
8

As an alternative to Marc's answer, suppose $GL_{n}(\mathbb{R})$ was cyclic, generated by the matrix $A$. Then every invertible matrix $B$ would be $A^{n}$ for some $n$. Since the determinant is multiplicative, we would have $\det B = (\det A)^n$. Since $\det B$ can take any value in $\mathbb{R} \setminus \{0\}$, consider what this would imply about whether $\mathbb{R} \setminus \{0\}$ is cyclic or not.

  • 0
    @user1729 : Indeed. Thanks for clarifying!2012-11-22