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I'm a bit stuck into this problem.

If $G$ is a group with order $364$, then it has a normal subgroup of order $13$.

I have tried to use Sylow III but all I could conclude was that the $7$-subgroup of Sylow in $G$ is normal and that the number of $13$-Subgroups of Sylow are $1$ or $14$. Some people told me that I can do this through "counting elements" but I don't know how I could do that. Can someone help me?

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Counting elements normally goes like this: if I had $14$ Sylow $13$-subgroups, then I'd have $14 \cdot 12 = 168$ distinct elements of order $13$. Normally you can then say that this number is too large, although nothing obvious pops out here. –  Dylan Moreland May 4 '12 at 20:40
    
I don't think counting elements is the best appraoch to this problem. Use the fact that the Sylow 7-subgroup is normal to deduce that it must be centralized by an element of order 13, and then use that fact to rule out the possibility of 14 Sylow 13-subgroups. –  Derek Holt May 4 '12 at 20:42
    
@DerekHolt, what do you mean by "it must be centralized by an element of order 13$? –  Marra May 4 '12 at 20:48
    
@Gustavo: say $P$ is the unique $7$-subgroup, and $x$ is a generator of $P$. If $y$ is an element of order $13$, then $x^y$ has to be a generator of $P$, and $x^{y^{13}}=1$; and $y$ acts as an automorphism on $P$. Since the cyclic group of order $7$ has automorphism group of order $6$, it contains no element other than the identity of order dividing $13$; so the action of $y$ must be the identity automorphism. That is, $x^y = x$, and so $py=yp$ for all $p\in P$. That is, $y$ centralizes $P$. –  Arturo Magidin May 4 '12 at 21:34
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2 Answers

up vote 10 down vote accepted

Suppose $G$ is a group of order $364 = 2^2 \cdot 7 \cdot 13$.

As you noted, there is a normal subgroup $P$ of order $7$ by Sylow's theorems. By Cauchy's theorem, there is a subgroup $K$ of order $13$ in $G$.

Now $P$ is normal, so $PK$ is a subgroup of order $7 \cdot 13$. Since $PK$ has index $4$, there exists a homomorphism $\phi: G \rightarrow S_4$ with $\operatorname{Ker}(\phi)$ contained in $PK$. You can see that this implies that $\operatorname{Ker}(\phi)$ equals $PK$ and thus $PK$ is a normal subgroup. Since every group of order $7 \cdot 13$ is cyclic and subgroups of a normal cyclic subgroup $PK$ are also normal in $G$, the subgroup $K$ is a normal subgroup.

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Actually, this is not enough. You need that subgroups of a cyclic group are characteristic. –  user641 May 5 '12 at 3:51
    
@Steve D I think this proof as it stands is correct. Maybe one should point out that every cyclic group has exactly one subgroup of each possible order. –  Olivier Bégassat May 5 '12 at 7:11
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@SteveD: What I meant is that if $T$ is a normal cyclic subgroup of $G$, then any subgroup of $T$ is normal in $G$. As you said, one way to prove this is to use the fact that subgroups of a cyclic group are characteristic. –  Mikko Korhonen May 5 '12 at 9:05
    
I'm going to try to show the missing parts of it, but thank you very much for the answer! –  Marra May 5 '12 at 13:08
    
@GustavoMarra: No problem. Since this is homework, I didn't want to give a complete solution. If you get stuck you can ask here in the comments and me or somebody else can help you. –  Mikko Korhonen May 5 '12 at 18:46
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After m.k.'s proof another one. There is indeed exactly one $7$-sylow subgroup, let's call is $S$: it is normal in $G$. There is also exactly one Sylow $13$-group in $G/S$ because the number divides $4$ and has residue $1$ modulo $13$. Take some generator of this unique Sylow $13$-subgroup of $G/S$. It has exactly $7$ antecedents in $G$. Also, any Sylow $13$-subgroup of $G$ intersects $S$ trivially so projects isomorphically onto the unique Sylow Subgroup of $G/S$. Each Sylow $13$-subgroup of $G$ thus contains one of the seven antecedents mentioned above, and two distinct Sylow $13$-subgroups of $G$ must intersect trivially, thus there areat most $7$ Sylow $13$-subgroups in $G$. But there can only be $1$ or $14$ so there is exactly one and it is normal.

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