7
$\begingroup$

This is part of a question from Hungerford's section on Sylow theorems, which is to show that any group with order 12, 28, 56, or 200 has a normal Sylow subgroup. I am just trying the case for $|G| = 12$ first.

I have read already that one can't conclude in general that $G$ will have a normal Sylow 2-subgroup or a normal Sylow 3-subgroup, so I am a bit confused on how to prove this. Here is my start, but it's not that far. Let $n_2, n_3$ denote the number of Sylow $2$- and $3$-subgroups, respectively. Then:

$n_2 \mid 3$ and $n_2 \equiv 1 (\operatorname{mod } 2)$, so $n_2 = 1$ or $3$. Similarly, $n_3 \mid 4$ and $n_3 \equiv 1 (\operatorname{mod } 3)$, so $n_3 = 1$ or $4$.

I was thinking to assume that $n_2 \ne 1$, and prove that in this case, $n_3$ must equal $1$. But I don't see how to do this. One could also do it the other way: if $n_3 \ne 1$, then show $n_2 = 1$.

$\endgroup$
  • $\begingroup$ tried counting ??? $\endgroup$ – user87543 Nov 3 '13 at 5:16
  • $\begingroup$ Have you already studied actions of groups on sets? In particular, the action of a group on the set of left cosets of one of its subgroups (the regular action)? $\endgroup$ – DonAntonio Nov 3 '13 at 5:16
  • $\begingroup$ We have covered a bit about group actions, but I'm not that good with them yet. Edit: counting worked very well for a proof by contradiction here, thanks! $\endgroup$ – user104986 Nov 3 '13 at 5:19
7
$\begingroup$

Hint: If $n_3 = 4$, then how many elements of order 3 are there in the group? How many elements does that leave for your groups of order 4?

$\endgroup$
  • $\begingroup$ Just got it, thanks! Hopefully I can do something similar for the larger orders. I'll try those ones now. $\endgroup$ – user104986 Nov 3 '13 at 5:25
4
$\begingroup$

Suppose $n_2 \neq 1$ and $n_3 \neq 1$. Then $n_2 = 3$ and $n_3=4$. Note that the intersection of any sylow 3-subgroup is trivial and the intersection of a sylow $2-$subgroup and a sylow $3-$subgroup is trivial. Also the intersection of any two sylow $2-$subgroup is a subgroup of size at most 2. The four sylow $3-$subgroup will contribute $4 \times (3-1)=8$ non-identity elements to the group. Also two of the sylow subgroup will contribute at least four non-identity elements to the group. So together with identity element, the group has at least 13 elements, which is not possible. Therefore either $n_2=1$ or $n_3=1$.

$\endgroup$
  • $\begingroup$ Sorry to bother you on an older post, but how exactly do you see that the Sylow 2-subgroups contribute at least four non-identity elements? If I am thinking about it correctly, each of the Sylow 2-subgroups has order $4$ (and they cannot intersect?) so we get $9$ non-identity elements from these groups. Is this correct? $\endgroup$ – Mike Pierce Nov 23 '15 at 5:44
  • $\begingroup$ @MikePierce Its alright. If you want to show that they cannot intersect(in this particular case), then i think you have to show. Otherwise, if you view them as just subgroups, then they may intersect at a subgroup of order 2. Generally, for any group $G$ and for any prime $p\mid G$, two sylow $p$-subgroups of $G$ may have a non-trivial intersection. As a simple counterexample, consider the subgroups $<r^3,s>$, $<r^3,rs>$ and $<r^3,r^2s>$ of $D_{12}$ $\endgroup$ – D. N. Nov 23 '15 at 6:29
  • $\begingroup$ Yeah, that was what I though could happen. Then how do we know that (at least) one of the Sylow 2-subgroups has order $4$? (that they don't all have order $2$?) $\endgroup$ – Mike Pierce Nov 23 '15 at 6:32
  • $\begingroup$ By definition of a sylow $p$-subgroup of a group $G$, they must be of order $p^n$, where $p^n\mid G$ and $p^{n+1}\not \mid G$. So in group of order 12, they must be of order 4. $\endgroup$ – D. N. Nov 23 '15 at 6:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.