# Is every group of order $21$ cyclic? [duplicate]

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 ?

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

• 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). – user1729 Apr 21 '13 at 14:55

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.

• thanks Your answer is really helpful to me – rst Apr 21 '13 at 16:43

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.

• In fact, there is not just "a", but "the" nontrivial homomorphism $\mathbb Z/3\to\operatorname{Aut}(\mathbb Z/7)$. – Hagen von Eitzen Apr 21 '13 at 14:32
• @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. – Pete L. Clark Apr 21 '13 at 14:47
• @Hagen, I meant to ping. – Pete L. Clark Apr 21 '13 at 15:09
• @PeteL.Clark Aw, right, I shot too fast – Hagen von Eitzen Apr 21 '13 at 17:55