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The following is taken from John D. Dixon, Problems in Group Theory, page 83. I have some questions concerning the solution.

Find the normalizer $N$ of the subgroup $\langle (1 2 3 \cdots n) \rangle$ in the symmetric subgroup $S_n$, ($n \ge 1$)

Solution: If $x \in N$, then $x^{-1}(1 2 \cdots n)x = (1^x 2^x \cdots n^x)$, where $x$ maps the letter $i$ onto $i^x$. Since $x^{-1}(1 2 \cdots n)x$ generates $\langle(1 2 \cdots n)\rangle$, it is equal to $(1 2 \ldots n)^k$ for some integer $k$ relatively prime to $n$. Thus $i^x + k \equiv (i+1)^x \pmod{n}$ for $i = 1,2,\ldots, n$. In particular, it follows that $i^x = k(i-1) + i^x \pmod{n}$. Thus $x$ is a permutation of the form $i^x = ki + l \pmod{n}$, where $k$ and $l$ are integers, $0 \le k, l < n$, with $k$ relatively prime to $n$. Conversely, any permutation of this form is in $N$.

  • Should "$i^x + k \equiv (i+1)^x \pmod{n}$ for $i = 1,2,\ldots, n$" be changed to: $i^x + k \equiv (i+1)^x \pmod{n}$ for $i = 1,2,\ldots, n-1$ and $n^x + k \equiv 1^x \pmod{n}$?

  • "[...] the form $i^x = ki + l \pmod{n}$" should by replaced by "[...] the form $i^x = k(i-1) + l$" and the following phrase should be added: "with $i = 2,\ldots, n$ and $1^x = l$ for $1 \le l \le n$" (instead of $0 \le l < n$)

Am I right, or is the original proof fine as it stands?

1 Answers 1

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Your first point is correct, though the only way $i^x+k\equiv (i+1)^x$ (mod $n$) makes sense for $i=n$ is to interpret it exactly how you rewrote it.

For the second point, yes the proof looks a little confused, so I will rewrite a couple of sentences adding an explanation of the piece you're asking about:

In particular, it follows that $i^x\equiv k(i-1)+1^x$ (mod $n$). Setting $0\le l< n$ such that $l\equiv 1^x-k$ (mod $n$) we have $i^x\equiv ki+l$ (mod $n$) where $k,l$ are integers with $0\le k,l