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This question already has an answer here:

solution :-

$21= 3 \times 7$

there is only one Sylow $3$ and Sylow $7$ subgroup

so, Sylow $3$ and Sylow $7$ subgroup are normal in group $G$

so $G$ is cyclic group of order $21$.

Am I right ?

somebody told me that group of order $21$ is not cyclic.

he gave me this link.

If group of order 21 is not cyclic, then can we understand it by Sylow method ?

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marked as duplicate by user26857, Brandon Carter, Dennis Gulko, Micah, Davide Giraudo Apr 21 '13 at 15:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ One of the answers in the question Metin Y. links to covers this (so although this question is not actually an exact duplicate, it is covered by the answers). $\endgroup$ – user1729 Apr 21 '13 at 14:55
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In a group of order 21,

The number of 3-sylow groups is equal to 1 mod 3 and divides $7$, so the possibilities are $1$ and $7$.

The number of 7-sylow groups is equal to 1 mod 7 and divides 3, so the possibilities is only $1$, there is a unique and hence normal 7-sylow group.

You have already dealt with the case of a unique 3-sylow group so let's consider if there were $7\quad 3$-sylow groups:


A 3-sylow group in this case is just $C_3$ we can see if any two 3-sylow groups overlap (except the identity) they are equal so we have 14+1 (the identity) elements from 3-sylow groups leaving $6$ elements over exactly what we need for the 7-sylow $C_7$. So it seems like at least one group like this will exist.

Now you can try to find a way to construct it but not using Sylow theory.

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  • $\begingroup$ thanks Your answer is really helpful to me $\endgroup$ – rst Apr 21 '13 at 16:43
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Actually there is a unique nonabelian group of order $21$ up to isomorphism. To get one, construct a nontrivial homomorphism $\mathbb{Z}/3 \to \operatorname{Aut}(\mathbb{Z}/7)$ and form the semidirect product.

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  • $\begingroup$ In fact, there is not just "a", but "the" nontrivial homomorphism $\mathbb Z/3\to\operatorname{Aut}(\mathbb Z/7)$. $\endgroup$ – Hagen von Eitzen Apr 21 '13 at 14:32
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    $\begingroup$ @Haegn: Since $\operatorname{Aut}(\mathbb{Z}/7\mathbb{Z})$ is cyclic of order $6$, it contains $\varphi(3) = 2$ elements of order $3$ -- explicitly, multiplication by $2$ and multiplication by $4$ -- so there are $2$ nontrivial homomorphisms from $\mathbb{Z}/3\mathbb{Z}$. They give rise to ismorphic semi-direct products though, as Serkan says. $\endgroup$ – Pete L. Clark Apr 21 '13 at 14:47
  • $\begingroup$ @Hagen, I meant to ping. $\endgroup$ – Pete L. Clark Apr 21 '13 at 15:09
  • $\begingroup$ @PeteL.Clark Aw, right, I shot too fast $\endgroup$ – Hagen von Eitzen Apr 21 '13 at 17:55

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