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I just need someone to check this argument.

Let $G$ be a nonabelian group of order $2009$. The prime factorization of $2009$ is $7^2 \cdot 41$. Let $n$ be the number of Sylow 7-subgroups.

Then $n \equiv 1$ mod $7$ and $n$ divides $41$. Since $41$ is prime, we must have $n =1$ or $n=41$, but $41 \equiv 6$ mod $7$. Thus $n=1$ and we have a unique Sylow 7-subgroup $H$.

Since $H$ is the unique Sylow 7-subgroup, we have that $H$ is a normal subgroup of $G$. Taking the quotient we have that $|G/H|=41$, so $G/H$ is cyclic of order 41.

Since $G/H$ is cyclic, $G$ is abelian. Thus there is no nonabelian group of order $2009$.

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    $\begingroup$ In $G = S_3$, you have $H = A_3$ normal and cyclic, and $G/H$ is also cyclic, but $G$ is not abelian. $\endgroup$ – Mikko Korhonen May 18 '14 at 16:29
  • $\begingroup$ Metacyclic (a cyclic extension of a cyclic group) need not imply cyclic. There is a correct version of this idea, which states that if $G/Z(G)$ is cyclic then it is trivial. $\endgroup$ – blue May 18 '14 at 16:48
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If $G/H$ is cyclic then you can not say $G$ is abelian as Mikko pointed out.

You should go on like that;

Let $q=41$ then $n_q$ must divides $49$ and $n_q\equiv 1 \ mod \ (41) \implies n_q=1 $.

Thus all sylow subgroup is normal, $G\cong H\times K$ where $|H|=49$ and $|K|=41$ since any group of order $p^2$ is abelian, $H$ is abelian then clearly $G$ is abelian.

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  • $\begingroup$ Alright thanks. Indeed I confused the result which states if $G/Z(G)$ is cyclic, then $G$ is abelian. I suppose that's what I get for chaining theorems and not thinking. What's the best way to see that if all sylow subgroups are normal then $G$ is the product of its sylow subgroups? $\endgroup$ – Frost Boss May 18 '14 at 17:02
  • $\begingroup$ You are welcome. $\endgroup$ – mesel May 18 '14 at 17:03
  • $\begingroup$ Let $P_1,P_2,...,P_n$ be all sylow subgroups of $G$ and all of them is normal. Since $P_i\cap P_j=e$ then $P_1.P_2...P_n \cong P_1\times P_2\times...\times P_n$ and when you compute the order of product you can see that it must be whole group. $\endgroup$ – mesel May 18 '14 at 17:11
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    $\begingroup$ Just to be finnicky, if you have normal subgroups $K_{i}$ of a group $G = K_{1} K_{2} \dots K_{n}$, and $K_{i} \cap K_{j} = \{ e \}$ for $i \ne j$, you cannot deduce that $G$ is isomorphic to the direct product of the $K_{i}$ - think the Klein four-group, with the $K_{i}$ being the three subgroups of order $2$. You need that each $K_{i}$ intersects trivially the product of the $K_{j}$ with $j \ne i$. $\endgroup$ – Andreas Caranti May 18 '14 at 18:59
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    $\begingroup$ @AndreasCaranti: I know, I wrote it to be clear, your comment was useful. $\endgroup$ – mesel May 19 '14 at 15:59
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I posted an erroneous remark in a (in the meantime deleted) answer - (thanks Mikko for pointing it out). Nevertheless, the arguments given by Mesel and Frost Boss nicely generalize to groups $G$ of order $p^2q$, with $p$, and $q$ primes, such that $p \lt q$, $p \nmid q-1$ and $q \nmid p^2-1$. Such a group has to be abelian.
(Note the conditions $p \lt q$, $p \nmid q-1$ imply $q \nmid p^2-1$, so $p \lt q$, $p \nmid q-1$ is sufficient to guarantee the abelianess).

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