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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.

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3 Answers 3

up vote 18 down vote accepted

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.

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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.

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In a sense, $GL_{n}(\mathbb{R})$ fails quite badly to be cyclic. We can throw away all matrix information but determinants and even then we can't recover much from a single generator... –  Isaac Solomon Nov 21 '12 at 15:13
Basically, you are using the fact that determinant is a surjective homomorphism $det: GL_n\mathbb{R}\rightarrow GL_1\mathbb{R}=(\mathbb{R}\setminus \{0\}, \ast)$ and $(\mathbb{R}\setminus \{0\}, \ast)$ is not cyclic. Just thought I'd make this clear... –  user1729 Nov 22 '12 at 14:42
@user1729 : Indeed. Thanks for clarifying! –  Isaac Solomon Nov 22 '12 at 17:22

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.

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...for $n>1$... –  user1729 Nov 22 '12 at 13:24
Thanks. I've mentioned that case now. –  Martin Argerami Nov 22 '12 at 14:41
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 $n=1$. –  user1729 Nov 22 '12 at 14:56

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