Any group of order $15$ is abelian(without sylow theorem) Prove that, any group of order $15$ is abelian (without help of Sylow's theorem or its application).
What I have done so far is, 
by class equation we know that $|G|=|Z|+\sum\frac{|G|}{C(a_i)}$. Now if I can show that $|G|=|Z|$ then the theorem is proved. Now order of $|Z|$ can not be $3$ or $5$, because if it is then $(G/Z)$ will be cyclic,which can not be since $G$ non-abelian.So, only possibilities are $|Z|=1$ or $15$. Now haw can I conclude that $|Z|\not=1$?
 A: If the center is trivial, the sum in the class equation equals $14.$ The centralizer of no element is the whole group, so it has to be a proper subroup, so each index is $3$ or $5.$  The only way to represent $14$ as a sum of $3$s and $5$s is $3 + 3 + 3 + 5 = 14.$ So, there are three conjugacy classes of elements of order $5,$ and one conjugacy class of elements of order $3,$ consisting of five elements, so there are five elements of order $3.$ But that's impossible, since the number of elements of order $3$ is even. (each subgroup of order $3$ has two nonzero elements, and the intersection of any two such is trivial).
A: Let, $G$ be a group of order $15$. Let, $Z$ be center if $G$. Then possible order of $Z$ are $1,3,5,15$. 
If $o(Z)=3$ or $5$ then $G$ is non-commutative and $(G/Z)$ must be non-cyclic. But, here $o(G/Z)=5$ or $3$ according as $o(Z)=3$ or $5$, which implies $(G/Z)$ is cyclic.Contradiction arise. Hence $o(Z)$ is not $5$ or $3$.
Now, by Cauchy's theorem, there is an element of order $5$. Let, $o(a)=5$ where $a\in G$. Let $<a>=H$(say). So, $H$ is only one subgroup of order $5$. Because if possible let $K$ be another subgroup of order $5$. Then $HK$ (may not be a subgroup) is a subset of G.
$$|HK|=\frac{|H||K|}{|H\cap K|}=\frac{5 *5 }{1}=25>15$$
$|H\cap K|=1$, since if there exists an non-identity element $b \in H \cap K$ then $H=K=<b>$. So, H is only one subgroup of order $5$.
Now, if possible let $|Z|=1$, $|G|=15(given)$, then from our class equation 
$$|G|=|Z|+\sum\frac{|G|}{|C(a_i)|}$$
$$\Rightarrow 14=\sum\frac{|G|}{|C(a_i)|}$$, also each of $\frac{|G|}{C(a_i)}$ devides $15$, i.e. $\frac{|G|}{|C(a_i|)}=3$ or $5$.
So, the only possible way to express $14$ is , $14=3+3+3+5$ $\Rightarrow \exists$ $a_1,a_2,a_3$ such that $\frac{|G|}{|C(a_1)|}=\frac{|G|}{|C(a_2)|}=\frac{|G|}{|C(a_3)|}=3$ $\Rightarrow |C(a_1)|=|C(a_2)|=|C(a_3)|=5$.
Now, since $H$ is the only subgroup of order $5$. $C(a_1)=C(a_2)=C(a_3)=H$ which is a contradiction, since $C(a_1),C(a_2),C(a_3)$ are same means $cl(a_1),cl(a_2),cl(a_3)$ are same which can not be true. Since they are different. Hence only possibility is $o(Z)=15 \Rightarrow G=Z$.
So, G is abelian. 
Now upto isomorphism there can be only two such abelian group $\mathbb{Z_3}$ X $\mathbb{Z_5}$ and $\mathbb{Z_{15}}$ since $g.c.d.(3,5)=1$, $\mathbb{Z_3}$ X $\mathbb{Z_5}$ and $\mathbb{Z_{15}}$ are isomorphic. Hence, $G$ is isomorphic to $\mathbb{Z_{15}}$. so, $G$ is cyclic.
