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S Apr 24 at 8:54 history suggested The Amplitwist CC BY-SA 4.0
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Mar 21, 2011 at 13:03 comment added Max Horn Olivier, you are absolutely right, my formula for the 2-Sylow as given is incorrect. Luckily, the final result is still correct, as it only depends on the the subgroup of all elements of order 2. This still has the shape $\mathbb{Z}/2\mathbb{Z})^{m+a}$, where $a=0,1,2$ depending on whether $k=0,1$ or $k=2$ or $k>2$.
Mar 18, 2011 at 21:24 comment added Olivier Bégassat For instance your formula yields that the $2$ Sylow of $\mathbb{Z}/17\mathbb{Z}$ should be $\mathbb{Z}/2\mathbb{Z}$ while it actually is $\mathbb{Z}/16\mathbb{Z}$.
Mar 18, 2011 at 21:21 comment added Olivier Bégassat so, I think the formula for the $2$ Sylow should be $(\mathbb{Z}/2^k\mathbb{Z})^{\times}\times\prod_{i=1}^r \mathbb{Z}/2^{\nu_i}\mathbb{Z}$ where $\nu_i$ is the highest exponent of $2$ in $P_i-1$. Correct me if I'm wrong, but I think this is the correct formula. I just noticed that I changed your notation, what you call $m$ I call $r$.
Mar 18, 2011 at 21:15 comment added Olivier Bégassat Hi Max, I think your $2$ Sylow formula might be wrong. I agree with what you've said before that. Set $n=2^k\cdot\prod_{i=1}^r p_i^{m_i}$ the decomposition into prime factors. Then $\mathrm{Aut}(N)\simeq (\mathbb{Z}/n\mathbb{Z})^{\times}\simeq (\mathbb{Z}/2^k\mathbb{Z})^{\times}\times\Prod_i (\mathbb{Z}/p_i^{m_i}\mathbb{Z})^{\times}$. The factor corresponding to the prime $2$ depends on wether $k\leq 2$, but is a $2$ group. For $p_i$ odd prime however $(\mathbb{Z}/p_i^{m_i}\mathbb{Z})^{\times}\simeq \mathbb{Z}/(p_i-1)\mathbb{Z}\times\mathbb{Z}/p_i^{m_i-1}\mathbb{Z}$.
Mar 18, 2011 at 12:26 history answered Max Horn CC BY-SA 2.5