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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 $n_2 = 1$.

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tried counting ??? – Praphulla Koushik Nov 3 '13 at 5:16
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)? – DonAntonio Nov 3 '13 at 5:16
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! – user104986 Nov 3 '13 at 5:19

2 Answers 2

up vote 5 down vote accepted

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?

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

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 $n_3=1$.

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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? – Mike Pierce Nov 23 at 5:44
@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 $D_{12}$ – D. N. Nov 23 at 6:29
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 $2$?) – Mike Pierce Nov 23 at 6:32
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. – D. N. Nov 23 at 6:36

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