I am reading the solution here: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2004&task=show_msg&msg=1066.0001
Assume G non-abelian. Case 2) all $4$-sylow subgroups of $G$ are cyclic. Take any element $b$ in $G$ of order $4$. Then since $<g>$ is normal in $G$, $bgb^{-1} = g^i$ for some $i$. $i \neq 1$ since $G$ is nonabelian. $i$ is not $0$ either (clearly!). if $i = 2$ then $b^3gb^{-3} = g^3$. If $i = 3$ then $bgb^{-1} = g^3$. Either case you get an element $b$ of order $4$ such that $bgb^{-1} = g^3$. So $G$ is generated by $a,b$ subject to $|a| = 5$, $|b| = 4$, $bgb^{-1} = g^3$. In the case $bgb^{-1} = g^4$ you get a different group.
So in the case of $i=4$, we have $bgb^{-1} = g^4$, so we have $b^3gb^{-3} = g^4 \neq g^3$, so for the case $i = 4$, it's different from $i = 2$ and $i = 3$, but $i = 2$ and $i = 3$ are the same. But if we use $b^2gb^{-2}$ instead of $b^3gb^{-3}$, then we get that for $i = 2$ that $b^2gb^{-2} = g^4$, and for $i = 3$ that $b^2gb^{-2} = g^4$, so it would imply that $i = 2$ and $i = 3$ are all the same as $i = 4$. Why isn't this true? Why do we specifically only care about the $b^3gb^{-3}$ case, and not about $b^2gb^{-2}$?
The subgroup generated by $b$ is acting on the subgroup generated by $g$ via conjugation (this works because $\langle g\rangle$ is normal in $G$). The reason why the $i = 2$ and $i = 3$ case are "the same" is because $b \mapsto b^3$ is an automorphism of $\langle b\rangle$, whereas $b \mapsto b^2$ is not ($b^2 = (b^3)^2$, but $b \neq b^3$). The automorphism $g \mapsto g^3$ of $\langle g\rangle$ induced by conjugation by $b$ or $b^3 = b^{-1}$ has order $4$ (it is an isomorphism of $\langle b\rangle$ with $\text{Aut}(\langle g\rangle)$, the automorphism $g \mapsto g^4$ induced by conjugation by $b$ only has order $2$, and represents a mere homomorphism of $\langle b\rangle$ into $\text{Aut}(\langle g\rangle)$.
Thus we have two distinct semi-direct products $\langle g\rangle \rtimes \langle b\rangle$, the first two (the $i = 2,3$ cases) are isomorphic.
In the $i = 4$ case, we have:
$b^2gb^{-2} = b(bgb^{-1})b^{-1} = bg^4b^{-1} = (bgb^{-1})^4 = (g^4)^4 = g^{16} = g$ (since $|g| = 5$).
Conjugation by $b^2$ is not equivalent to conjugation by $b$.