Simple method the show that a group of order 15 is cyclic How can i show with quite simple methods that a group of order 15 is cyclic? 
 A: Here's another method: there is a theorem that you may have seen before that states

If $H \le G$ is a subgroup such that $p=[G:H]$ is the smallest prime dividing $|G|$, then $H$ is normal.

If you apply this theorem to the subgroup generated by an element of order 5 in $G$ when $|G| = 15$, you see that it is normal.  Let $N$ be a subgroup of order 5, and $H$ a subgroup of order 3.  Then $H$ acts by conjugation on $N$, since $N$ is normal.  This gives a homomorphism $\varphi: H \to Aut(N)$ where $Aut(N) \cong \mathbb{Z}/4\mathbb{Z}$.  This automorphism group has no elements of order 3, so $\varphi$ must map $H$ to the trivial automorphism on $N$.  Stated differently, this means that $hnh^{-1} = n$ for $h \in H$, $n \in N$, so the subgroups $H$ and $N$ commute with each other!.
From there, you can see that $G$ is generated by the element $hn$.
A: By Cauchy's theorem,we have there exists $a$ and $b$ in $G$ with orders $3$ and $5$. Let $A = \langle a\rangle$ and $B=\langle b \rangle$ . They are Sylow's group and also are Normal.
Now $ab$ is not in either $A$ or $B$. Hence order of $ab$ can't be $1,3,5$. Hence order of $ab$ is $15$. And $ab$ generates the group.
EDIT:
The groups $A$ is unique because $15 = 3 \times 5$ and $n_3 |5$ & $n_3 = 1+3k$,hence $n_3 = 1$ where $n_3$ is no of Sylow 3 subgroups.(we are using Sylow's third theorem)
Now uniqueness of Sylow's p-group implies normality.(Sylow's second theorem)
Order of $ab$ is quite clear I think.
Alternatively,as $A$ and $B$ are normal subgroups intersecting trivially,$G=AB=A \times B \simeq \mathbb Z_3 \times \mathbb Z_5 \simeq Z_{15}$
Hope its enough.
Note:For $|G| =21$ ,$n_3 |7$ and $n_3=1+3k$ may have solution $n_3=6$ Hence number of sylow 3 subgroups may not be 1.Now if we are given that it has only one subgroup of order 3 and only one subgroup of order 7,then indeed the group will be cyclic.
A: Assume that $G$ is not cyclic.
Then $G=A\cup B\cup\{1\}$, where $A$ is the set  of elements of order 5 and $B$ is the set of elements of order 3.
We know that $|A|$ is a multiple of 4 because two different cyclic subgroups of  prime order have at most the neutral element in common.
Similarly, $|B|$ is a multiple of 2.
Also, we know that $|A|>0$ and $|B|>0$.
Let $a$ be an element of order 3, $b$ an element of order 5.
The centralizer of $\langle a\rangle$ must be either $\langle a\rangle$ or $G$. In the latter case, we have especially $ab=ba$ and one verifies that $(ab)^3=b^3\ne1$ and $(ab)^5=a^5\ne1$ and $(ab)^{15}=1$, whence $G=\langle ab\rangle$.
Therefore we may assume that the centralizer of $\langle a\rangle $ is $\langle a\rangle $ itself.
The group $\langle a\rangle$ operates on $A$ by conjugation. The only fixed points of this operation are the elements of the centralizer, but they are not of order 5. Hence the operation has no fixed points and we see that  $|A|$ is a multiple of 3.
By the same argument with roles of 3 and 5 swapped, we find that $|B|$ is a multiple of 5.
We conclude that $|A|$ is a positive multiple of 12 and $|B|$ a positive multiple of 10, hence $|G|\ge 12+10+1=23$, contradiction.

Remark: If we tried the same with order 21 instead of 15, we would find (mutatis mutandis) that $|A|$ is a multiple of 6(!) and $|B|$ a multiple of $14$, so that $|G|\ge 6+14+1$ does not produce a contradiction. And indeed, there is a non-cyclic group of order 21 (with necessarily 6 elements of order 7 and 14 elements of order 3), obtained as a semidirect product by letting $C_3$ operate non-trivially on $C_7$.
