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I'm working my way through Stillwell's Naive Lie Theory, and feel like I'm missing something simple with problem 7.4.1, which reads:

Show that each $A \in N_\delta (\mathbf 1)$ has a unique nth root for $n=1,2,3,...$.

Here, $N_\delta (\mathbf 1)$ is a neighborhood of the identify element of a Lie group $G$ that is mapped into the tangent space of $G$ by $\log$. I assume he means that there is a unique root in $N_\delta (\mathbf 1)$

But if we just consider $U(1)$, it seems like whatever neighborhood of $\mathbf 1$ I choose, for large enough $n$ there will be multiple roots of unity in the neighborhood. What am I missing here?

Edit:

Based on the discussion with @Alp, it does seem like the phrasing of the question is a bit off. Here's a rewrite that seems to capture the intent:

Show that for each $n\in \mathbb Z^+$ there exists a neighborhood $N_{\delta_n}(\mathbf 1)$ such that each $A\in N_{\delta_n}(\mathbf 1)$ has a unique $n$th root in $N_{\delta_n}(\mathbf 1)$.

Isaac
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  • It seems $\delta$ is meant to work for one $n$ (or finitely many $n$'s) at a time. – Alp Uzman May 02 '22 at 07:21
  • That's a thought. Maybe he also just didn't mean to include "unique". – Isaac May 02 '22 at 13:45
  • I think uniqueness is the important point in this exercise, considering the next exercise in the book, but you may be right. – Alp Uzman May 02 '22 at 15:25
  • That's true. My rephrasing, then, is: Show that for each $n\in \mathbb Z^+$ there is a neighborhood $N_n(\mathbf 1)$ such that each $A\in N_n$ has a unique $n$th root in $N_n(\mathbf 1)$. – Isaac May 02 '22 at 16:42
  • It might be better to write $\forall n, \exists \delta_n: N_{\delta_n}(1)$... to be in accordance with the book's notation. – Alp Uzman May 02 '22 at 16:45
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    Fair. And thanks! – Isaac May 02 '22 at 17:16

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