How to show that group of order 15 is cyclic? Let $G$ be a group of order $15$ then prove that it is cyclic.
By Lagrange's theorem, its non identity elements can only have order $3$ or $5$. Not all non identity elements are of order $\displaystyle 5$ because $\displaystyle \phi ( 5) =4$ does not divide $14$.
If all non identity elements are of order $\displaystyle 3$ then $G$ is of the form:
$\displaystyle G=\left\{1,a_{1} ,a_{1}^{2} ,a_{2} ,a_{2}^{2} ,\cdots ,a_{7} ,a_{7}^{2}\right\}$
$\displaystyle a_{1} a_{2} \in \left\{a_{i}^{r} :\ 3\leq i\leq 7,\ 1\leq r\leq 2\right\}$
Case 1: WLOG $\displaystyle a_{1} a_{2} =a_{3}$
By $\displaystyle a_{3}^{2} =a_{3}^{-1}$, we have $\displaystyle a_{1} a_{2} a_{1} a_{2} =a_{2}^{2} a_{1}^{2}$
How do I get a contradiction here?
Case 2: WLOG $\displaystyle a_{1} a_{2} =a_{3}^{2}$
How do I get a contradiction here?
If I establish that there must exist at least one order 5 element (say b) then,
Clearly, $\displaystyle |\langle a \rangle \cap \langle b \rangle|=1$
Cosets of $\displaystyle \langle a \rangle $ are $\displaystyle \langle a \rangle, b\langle a \rangle, b^{2} \langle a \rangle,\ b^{3} \langle a \rangle, b^{4} \langle a \rangle $
Suppose $\displaystyle G$ is not cyclic, then for any $\displaystyle g\in G,\ |g|\neq 15$.
$\displaystyle  |G/Z( G) |=\frac{15}{|Z( G) |} \in \{15,5,3,1\}$
From here, I plan to show that $\displaystyle |Z( G) |=15$ so that $\displaystyle G=Z( G) \Longrightarrow G$ is abelian.
So I must reject the following:
$\displaystyle Z( G) \in \{1,3,5\}$
I have no idea how to do this either and even if I somehow manage to do this, I have no idea how to proceed further to show $G$ is cyclic.
I'll appreciate any hint in this. I would also like to mention that following are not allowed:
 1. Cauchy's theorem 


*Sylow's theorems


*Group actions
Lagrange's theorem, Cosets, External direct products, Internal direct products, factor groups (quotient groups) etc. are allowed.
 A: Note that $G/Z(G)$ cannot have a prime order. If it did, then $G/Z(G)$ would be a cyclic group, making $G$ abelian, making $G=Z(G)$ so that $|G/Z(G)|=1$ which is a contradiction. So, the only possible order of $Z(G)$ are $1$ and $15$. If $|Z(G)|=15$, then we are done as $Z(G)=G$. Now, suppose $|Z(G)|=1$. Then, use the class equation. We note that $|\mathcal{K}(x)|=1$ iff $x \in Z(G)$. Since $|Z(G)|=1$, every non-identity element $g$, is such that $|\mathcal{K}(g)| \in \{3,5\}$. So, $|G|=|Z(G)|+\displaystyle\sum|\mathcal{K}(g)|$. Since $|\mathcal{K}(g)| \in \{3,5\}$, $\displaystyle\sum\mathcal{K}(g)$ is a linear combination of $3$ and $5$ i.e $\displaystyle\sum\mathcal{K}(g)=3a+5b$ for some $a,b \in \mathbb{N} \cup \{0\}$. So, $15=1+3a+5b$, which is clearly a contradiction. Hence, $|Z(G)| \neq 1 \implies |Z(G)|=15$.
EDIT: Answer without using class equation. Every element in $G$ will have order $1,3,5,15$ from Lagrange's theorem. If there is some element of order $15$ then we are done. Suppose not. From Cauchy's theorem, there is an element of order 3, and there is an element of order $5$. Let $|p|=3$ and $|q|=5$. Let $P=<p>$ and $Q=<q>$. It is clear that $P \cap Q \leq P$ and $P \cap Q \leq Q$ hence, $|P \cap Q| \mid |P|,|Q| \implies |P \cap Q|=1$. Consider $PQ$. It has order $|PQ|=\frac{|P||Q|}{|P \cap Q|}=3\cdot 5=|G|$. Hence, $G=PQ$. We know that $AB$ is a subgroup of $G$, iff $AB=BA$ where $A,B$ are subgroups of $G$. Hence, $PQ=QP$, $PQP^{-1}=Q$. It is obvious that if $a \in Q$, then $a^{-1}Qa=Q$. The other elements of $G$ not in $Q$ will be in $P$ and from above $PQP^{-1}=Q$. Thus, $g^{-1}Qg=Q, \forall g \in G$, so that $Q \triangleleft G$ and similarly, $P \triangleleft G$. Now, we have $P,Q$ are abelian because they are cyclic and $P \cap Q=\{e\}$. Let $a \in P$ and $b \in Q$. Then $a(ba^{-1}b^{-1})=(aba^{-1})b^{-1} \in P,Q \implies aba^{-1}b^{-1} \in P \cap Q=\{e\}$, so $ab=ba$. Thus, for any $g(=p_1q_1), h(=p_2q_2) \in G$, $gh=p_1q_1p_2q_2=p_2q_2p_1q_1=hg$, so $G$ is abelian, hence $|pq|=15 \implies G=<pq>$.
