Classification of groups of order 30

Let $G$ be a group of order 30 and let $n_p$ be the number of its $p$-Sylow subgroups. By Sylow, $n_3\in\{1,10\}$ and $n_5\in\{1,6\}$. It is not possible that $n_3 = 10$ and in the same time $n_5 = 6$, since otherwise, there would be 20 elements of order 3 and 24 elements of order 5, contradicting the fact that $G$ contains only 30 elements.

So either the $3$-Sylow subgroup $P_3$ or the $5$-Sylow subgroup $P_5$ is normal, implying that $N = P_3 P_5$ is a subgroup of $G$. Because of $\gcd(3,5) = 1$, $P_3\cap P_5 = \{1\}$ and therefore $\lvert N\rvert = \lvert P_3\rvert\cdot\lvert P_5\rvert = 15$. By the classification of $pq$-groups and $5\not\equiv 1\bmod 3$, $N$ is cyclic. Since its index in $G$ is $2$, it is a normal subgroup. Any $2$-Sylow subgroup is a complement of $N$ in $G$.

So $G \cong \mathbb Z/15\mathbb Z \rtimes_{\varphi} \mathbb Z/2\mathbb Z$, where $\varphi : \mathbb Z/2\mathbb Z \to \operatorname{Aut}(\mathbb Z/15\mathbb Z)$ is a group homomorphism. We have $\operatorname{Aut}(\mathbb Z/15\mathbb Z) \cong (\mathbb Z/15\mathbb Z)^\times \cong (\mathbb Z/3\mathbb Z \times \mathbb Z/5\mathbb Z)^\times \cong \mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$. Since $\varphi$ is a homomorphism and the order of $1$ in $\mathbb Z/2\mathbb Z$ is $2$, the order of its image $\varphi(1)$ divides $2$. In $\mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$ there are four possible images, namely $\varphi(1)\in\{(0,0), (1,0), (0,2), (1,2)\}$. Since $\varphi$ is determined by $\varphi(1)$, this shows that there are at most $4$ isomorphism types of groups of order $30$.

Now look at the following four groups or order $30$: $$ \mathbb Z/30\mathbb Z,\quad D_{15},\quad \mathbb Z/5\mathbb Z\times S_3,\quad \mathbb Z/3\mathbb Z\times D_5 $$ It is not hard to check that they are pairwise non-isomorphic. So there are exactly these $4$ isomorphism types of groups of order $30$.

You should already have seen that a group of order 15 is cyclic. This follows readily from Sylow's theorems.

You say

I have found all the sylow subgroups and have also found that either sylow-3 or sylow 5 subgroup is Normal.

Suppose there is a normal 5-Sylow subgroup $F$, and let $T$ be a 3-Sylow subgroup. Then $F T$ is a subgroup of order 15, so it is cyclic. Moreover $FT$ has index 2, so it is normal. It follows that $T$ is also normal in $G$. A similar argument applies with 3 and 5 exchanged, so now you know that both a 3- and a 5-Sylow subgroup is normal.

Thus $G$ has a cyclic normal subgroup $FT$ of order 15. The last thing to do is to see how an element $\alpha$ of order 2 can act on $FT$. Consider separately the action of $\alpha$ on $F$ and $T$.