A group $G$ (possibly non-abelian) is divisible when for all $k\in \Bbb N$ and $g\in G$ there exists $h\in G$ such that $g=h^k.$ Is the group $\mathrm{GL}_n(\mathbb C)$ divisible? Or more precisely, for which $n$ is it divisible? (It clearly is for $n=0$ and $n=1$.)
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1When a matrix is diagonalizable, you can find $k$th roots for all $k$. So the question is for left non-diagonalizable matrices. – Thomas Andrews Mar 30 '13 at 00:20
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1The assertion is true for any diagonizable matrix, so you should look at non diagonalizable matrices. – Pedro Mar 30 '13 at 00:20
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@Bartek, Instead of changing the original question, why not make a new question? I'll upvote that one, too. – Stephen Mar 30 '13 at 00:30
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1@Steve OK, but should I ask two new questions or one? The two added questions are probably just as similar to each other as they are to the original question... – Bartek Mar 30 '13 at 00:32
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@Bartek, I would make two separate questions---I am not an expert, but it feels to me that most of the complication lies in allowing positive characteristic, and that once this is sorted out, the infinite matrices bit won't be too hard... so maybe just start by asking the same question for an arbitrary a.c. field. – Stephen Mar 30 '13 at 00:34
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@Steve OK, I'll do that now and wait with the infinite matrices. – Bartek Mar 30 '13 at 00:37
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Yes. The easiest way to see it is to observe that the exponential map sending an arbitrary (not necessarily invertible) matrix $A$ to $e^A$ is onto for $\mathrm{GL}_n(\mathbb{C})$. (This fails for some of your other favorites matrix groups). Then $$(e^{\frac{1}{n}A})^n=e^A$$ gives you $n$th roots.
Stephen
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Thank you very much. Could you explain why the map is onto? I'm not very comfortable manipulating the matrix $\exp.$ – Bartek Mar 30 '13 at 00:36
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@Alex and Steve: Yes, of course you are both right. (I can't be the only person who continually assumes all manifolds are connected, can I?) – Jason DeVito - on hiatus Mar 30 '13 at 00:39
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@JasonDeVito Unless you care about dimension $0$ Lie groups--I heard those can be pretty important :) – Alex Youcis Mar 30 '13 at 00:40
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1@Bartek, Using Jordan decomposition, write an arbitrary invertible matrix as a product $g=s u$ of a semisimple matrix $s$ and a unipotent matrix $u$ that commute with one another. Then take logarithms separately. – Stephen Mar 30 '13 at 00:41
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@Steve I haven't heard of the Jordan decomposition before. Could you add a short explanation? (If it's too much trouble, then it's not a problem. I'll read up -- there's a Wikipedia article about it and I see there are references there.) – Bartek Mar 30 '13 at 01:23
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@Bartek, It's kind of a long story. Probably wikipedia will sort you out! – Stephen Mar 30 '13 at 01:28
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Thank you. I will have to think about this. I'll comment if I have problems. – Bartek Mar 30 '13 at 01:40
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@JasonDeVito: Can we say that the divisibility of $GL_n(\mathbb C)$ is being inherited from divisibility of $\mathbb C$? – Mikasa Mar 30 '13 at 09:31
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@Babak: There may some sense in which what you're saying is true, but it's not immediately obvious to me. – Jason DeVito - on hiatus Mar 30 '13 at 14:17
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Hint: The following fact is from to J.J.Rotman and I think you can satisfy it for $G$:
If $A$ is a group, $m\in\mathbb Z$, and $m_A:~A\to A$ by $$a\longmapsto ma$$ then $A$ is divisible iff $$\forall m\neq 0,~~m_A~~\text{is a surjection}$$
Mikasa
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