Let $H= \langle a,b \rangle$ be the subgroup of $S_5$ generated by the two permutations: $a=(1\,4\,5\,2)$ and $b=(1\,3\,4\,2)$
$1.$ What is the order of the permutation $g=ab$?
$2.$ How many conjugates of $g$ are there in $S_5$?
$3.$ Prove that $\langle g \rangle$ is a normal subgroup of $H$.
$4.$ Prove that the subgroups $\langle a \rangle$ and $\langle b \rangle$ are conjugated in $H$.
My attempts
$1.$
I calculate $ab$:
$ab=(1\,4\,5\,2)(1\,3\,4\,2)=(1\,2\,3\,4\,5)$
Since $ab$ is a $5$-cycle, its order is exactly 5.
$2.$
Let $c \in S_5$. The permutations $c$ and $g$ are conjugated if and only if they have the same cycle type, that is if and only if $c$ is a $5$-cycle. This means that we must have:
$c=(c_1\,c_2\,c_3\,c_4\,c_5)$, with $c_i \in \{1,2,3,4,5\}$ for each $i=1,2,3,4,5$, with no repetitions.
For example, the permutation $(1\,3\,2\,5\,4)$ is acceptable, but $(2\,3\,4\,5\,3)$ is not. This makes me think that, chosen the first number, the second yields $4$ choices, the third $3$ choices, the fourth $2$ and the fifth $1$. This would make for $4!$ possible permutations, but I am not sure about this, since I do not really know how to exclude the fact that the first number can actually be chosen in $5$ different ways, which would make for $5!$ conjugates. But $|S_5|=5!$.
$3.$
Since $|g|=5$, it follows that $\langle g \rangle =\{g^p\,|\,p \in \{1,2,3,4,5\} \}$. The permutations $a$,$b$ are $4$-cycles, so they have order $4$ and $\langle a,b \rangle = \{a^tb^s\,|\,t,s\in\{1,2,3,4\} \}$
The subgroup $\langle g \rangle$ is normal in $H$ if and only if:
For each $g^p \in \langle g \rangle$ and $a^tb^s \in H$ for $p \in \{1,2,3,4,5\}$ and $t,s \in \{1,2,3,4\}$ , we have that
$(a^tb^s)^{-1}g^p(a^tb^s)=g^p$
I observe that $g^p=(1\,(1+p)\,(2+p)\,(3+p)\,(4+p))$, which is simply a consequence of the general properties of powers of $k$-cycles. Using the fact that, given $w \in S_n$ a $k$-cycle and $u \in S_n$ a permutation, we have that:
If $w=(w_1\,w_2\, ... \,w_k)$, then $u^{-1}wu=(w_1u\,w_2u\, ... \,w_ku)$.
I can write
$(a^tb^s)^{-1}g^p(a^tb^s)=(1a^tb^s\,(1+p)a^tb^s\,(2+p)a^tb^s\,(3+p)a^tb^s\,(4+p)a^tb^s)$
In order to conclude, I should be able to prove that the permutations $a^tb^s$ fix the numbers $1,(1+p),(2+p),(3+p),(4+p)$, which I do not know how to do.
I would be happy if you could help me with points $3.$ and $4.$, and of course, if I did any mistakes with the first two, please correct me.