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Let $2n = 2^a k$ for $k$ odd. Prove that the number of Sylow $2$-subgroups of $D_{2n}$ is $k$.

I managed to prove this result by showing that the normalizer of any Sylow $2$-subgroup is itself. The result immediately follows, in that case.

However, my problem is that I came across a different solution to this problem, and it appeals to me because it (claims to) enumerate all subgroups of $D_{2n}$ of order $2^a$.

The claim is that you can construct all the Sylow $2$-subgroups of $D_{2n}$ as follows: Let $j \in \{0, 1, \dots, k-1\}$ and let $t \in \{0, 1, \dots 2^{a-1}\}$. Then $A_{j} = \{r^{tk}, sr^{tk + j}\}$ is an enumeration of all subgroups of order $2^a$. That is, for each fixed $j$, we iterate over the values of $t$.

My problem is that, while this construction gives $k$ distinct subgroups of the desired order, they all have the element $r^k$ in common. But is it not the case that distinct Sylow $p$-subgroups intersect trivially?

combinator
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  • Well, this is an example that shows that your statement is simply not true. I suppose your idea comes from the prime case that is used for enumeration. – Phira Dec 29 '13 at 00:36
  • Do you mean to say that the result is not true or that the proposed solution is false? – combinator Dec 29 '13 at 00:38
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    I wanted to say that it is not the case that distinct Sylow $p$-groups intersect trivially. – Phira Dec 29 '13 at 00:52
  • Oh, okay. I was under the impression that Sylow subgroups of prime order must have trivial intersection. Is there anything concrete that can be said about when Sylow subgroups must intersect trivially? – combinator Dec 29 '13 at 01:00
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    @combinator Sylow subgroups of prime order have trivial intersection, yes. But the 2-Sylow subgroups here don't have prime order. In general, having trivial intersections of p-Sylow subgroups is a strong condition—at least if there are many p-Sylow subgroups—since it means that there will be lots of elements with order a power of p. – Andrew Dudzik Dec 29 '13 at 01:15
  • Can u please tell me why the normalizer of any Sylow 2 -subgroup is itself? – Ri-Li Sep 14 '14 at 07:22

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