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It seems that its explained pretty well online how to find a p-sylow subgroup, but Im having a hard time finding an explanation of how to find a p-sylow subgroups normalizer.

Take a specific 2-sylow subgroup of $S_4$: It will have order $2^3=8$ and could look something like $D_4$ in $S_4$ (symmetries of square): {e,(1234),(1432),(12)(34),(14)(23)...}. What is the best method for finding the normalizer of this group?

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  • $\begingroup$ "could look something like $D_4$..." In fact, it must be an isomorphic copy of $D_4$. If we label the vertices of the square with the numbers $1,2,3,4$, then $D_4$ permutes these numbers, so it is a subgroup of $S_4$. Moreover, it is a $2$-sylow subgroup of $S_4$ since it has the right order, and all of the $2$-sylow subgroups are conjugate, hence isomorphic. So every $2$-sylow subgroup is a copy of $D_4$, and by renumbering the vertices of the square, we can get distinct copies of $D_4$ in $S_4$. In particular, there is more than one $2$-sylow subgroup. Therefore...? $\endgroup$
    – user169852
    Dec 17, 2015 at 0:25
  • $\begingroup$ Sorry, I meant to say its an isomorphic copy (I wasn't using careful wording) $\endgroup$ Dec 17, 2015 at 0:47
  • $\begingroup$ It is also useful equality $N_G(P)=N_G(N_G(P))$ $\endgroup$
    – mesel
    Dec 17, 2015 at 5:42

2 Answers 2

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Addressing one of the additional cases raised in the comments: let us find the normalizer of a $5$-Sylow subgroup of $S_5$.

First, let $n_5$ denote the number of $5$-Sylow subgroups of $S_5$. From Sylow's theorem, we have $n_5 \equiv 1$ (mod $5$) and $n_5$ must divide $4! = 24$. The possibilities are $n_5 = 1,6$. In this case, a $5$-Sylow subgroup has order $5$, so it is generated by any $5$-cycle. Since there are more than four $5$-cycles, there must be more than one $5$-Sylow subgroup, so $n_5 = 6$.

Now we can use the fact that $|G:N_G(S)| = n_5 = 6$, where $S$ is any $5$-Sylow subgroup. So the normalizer of $S$ has index $6$, hence order $20$. Certainly $S \leq N_G(S)$. What else is in $N_G(S)$? Without loss of generality (relabeling if necessary), we may assume that $S = \langle(12345)\rangle$.

Note that there are four generators of $S$, namely: $$(12345)$$ $$(12345)^2 = (13524)$$ $$(12345)^3 = (14253)$$ $$(12345)^4 = (15432)$$ Any element $g \in S_5$ for which $g(12345)g^{-1}$ is one of the generators of $S$ will normalize $S$. We can easily see that, for instance, $$g(12345)g^{-1} = (13524)\quad\text{ if }\quad g = (2354)$$ This shows that $N_G(S)$ contains the cyclic subgroup $H = \langle(2354)\rangle$. Therefore, $N_G(S)$ contains $SH$. Moreover, $|SH| = |S||H|/|S \cap H| = (5)(4)/(1) = 20$, so $N_G(S) = SH$.

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  • $\begingroup$ This is good! Thanks Bungo! $\endgroup$ Dec 17, 2015 at 1:45
  • $\begingroup$ @ThePhysicsStudent: I'm glad it was helpful. $\endgroup$
    – user169852
    Dec 17, 2015 at 1:50
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    $\begingroup$ Well, it certainly divides $5! = 120$, but we can be more specific. Recall that if $G$ is any finite group and $H$ is a $p$-Sylow subgroup of $G$, then $n_p = |G:N_G(H)|$. Since $H \leq N_G(H)$, we see that $n_p$ must divide $|G:N_G(H)||N_G(H):H| = |G:H|$. In our case, $G$ is $S_5$, which has order $5!$, and $H$ has order $5$, so $|G:H| = |G|/|H| = 5!/5 = 4! = 24$. $\endgroup$
    – user169852
    Dec 17, 2015 at 2:25
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    $\begingroup$ @1ENİGMA1 Yes, we can. Note that $\langle a,b \rangle$ contains $a$ and $b$, hence it contains $\langle a \rangle$ and $\langle b \rangle$, hence it contains $\langle a \rangle \langle b \rangle$. For the reverse containment, note that $\langle a \rangle \langle b \rangle$ is a subgroup (as shown above), which of course contains $a$ and $b$. Since $\langle a ,b \rangle$ is the smallest subgroup containing $a$ and $b$ (i.e. it is contained in all other subgroups which contain $a$ and $b$), we must have $\langle a ,b \rangle \leq \langle a \rangle \langle b \rangle$. $\endgroup$
    – user169852
    Jan 4, 2018 at 16:55
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    $\begingroup$ @1ENİGMA1 That is a sufficient but not necessary condition. It is possible for $HK$ to be a subgroup of a group $G$ even if neither $H$ nor $K$ is a normal subgroup of $G$. For example, let $G = S_4$, and let $H$ be one of the $2$-sylow subgroups (order $8$) and $K$ be one of the $3$-sylow subgroups (order $3$). Neither $H$ nor $K$ is normal, but the product $HK$ has order $24$, hence is all of $S_4$, so it's certainly a group. In the specific example above, we know that $\langle a \rangle \langle b \rangle$ is a subgroup because we showed that it equals $N_G(H)$. $\endgroup$
    – user169852
    Jan 4, 2018 at 18:31
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A $2$-Sylow subgroup in $S_4$ is a maximal subgroup, meaning there is no other proper subgroup of $S_4$ containing it (this can be seen by the fact that it has prime index, namely $3$). Thus it is either its own normalizer or normal. Which is it?

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  • $\begingroup$ Thanks for the response Matt, but Im wondering about more generally? what if I had say a 5-sylow subgroup in $S_5$ ? or a 3-sylow in $S_5$ (something that isn't maximal) ? $\endgroup$ Dec 17, 2015 at 0:37
  • $\begingroup$ @Physics there exist methods that work in more than just one case, but in general you should expect every case to be special. This is true in most things; you do have tools, but most of the time they're only useful as background and something new is needed every time. $\endgroup$ Dec 17, 2015 at 1:04
  • $\begingroup$ ahh ok, thanks Matt! $\endgroup$ Dec 17, 2015 at 1:45

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