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

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.

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

• 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). 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.

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 $\mathbb Z/3\to\operatorname{Aut}(\mathbb Z/7)$. 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. Apr 21 '13 at 14:47