Determining which group of order 30 $G$ is given a relation of its elements. Let's say we have a group $G$ where $|G|=30$. I am familiar with the traditional argument using Sylow's theorems that gives rise to the fact that $G$ is isomorphic to one of the following groups:
$$\mathbb{Z}_{30},\mathbb{Z}_3\times D_5, \mathbb{Z}_5\times D_3, D_{15}$$
Obviously these groups are not isomorphic to one another since they have different numbers of elements of certain orders. We are given the additional information that $a,b\in G$ such that $|a|=2,|b|=15$, $ba=ab^4$. I am told that in this case $G\cong \mathbb{Z}_3\times D_5$; yet, I am not quite sure why this is the case.
If we write $ba=ab^4 \implies baa=ab^4a \implies b=ab^4a\implies b=ab^4a^{-1}$. Since $b^4$ and $b$ are conjugate in $G$, we have that $|b^4|=|b|=15$. Not quite sure where I'm going with this approach, though. A hint would be helpful!

The order of $b^4$ is the same as that of $b$ without reference to $a,$ since $4$ is relatively prime to $15.$ However, notice that the fact that 
$$b^4 = a b a^{-1}$$ implies (by raising both sides to fifth power) that $$b^5 = a b^5 a^{-1}.$$ Which means that $b^5$ is central. Further, $b^5$ generates the Sylow $3$-subgroup, which means there is exactly one such. Consider now $a$ and $b^3.$ Note that $ab^3 a^-1 = b^{12} = b^{-3},$ so $a$ and $b^3$ generate a $D_5.$ I claim now that $G$ is the direct product of the $C_3$ and the $D_5.$ Indeed, the intersection of the two of them has to be trivial (by order considerations), and pairs of elements commute (since one of them is central).
