This is a homework problem but I have stuck at some other chapter earlier so I am now completely lost what I am supposed to do. Clues and hints and suggestions of theorems that I should look up would be very useful.
Let $H,N$ be subgroups of $S_{5}$ such that: 1. $N$ is a normal subgroup in $H$. 2. the cycle $(i,j,k)$ is in $H$ for all distinct $1\leq i,j,k\leq 5$. 3. $H/N$ is Abelian. Show that $N$ contains all 3-cycles, i.e: prove that $(i,j,k)\in N$ for all distinct $1\leq i,j,k\leq 5$.
It's not that hard to show that $A_n$ (for $n \geq 3$) is generated by the $3$-cycles:
Even permutations can be decomposed into pairs of transpositions, if the pair is two identical transpositions, we can replace it by $e$ without affecting the product (this still leaves an even number of transpositions left).
If the transpositions are disjoint, we can realize them as the product of a pair of $3$-cycles:
$(a\ b)(c\ d) = (a\ b\ c)(b\ c\ d)$.
If the transpositions share one letter in common, they are a $3$-cycle:
$(a\ b)(a\ c) = (a\ c\ b)$.
So... the problem statement tells us $H = A_5$ or $H = S_5$.
We also know $H/N$ is abelian. This means $[H,H]$ (the subgroup generated by all commutators $[a,b] = aba^{-1}b^{1}$ for $a,b \in H$) is contained in $N$, because:
$NaNb = NbNa$ (since $H/N$ is abelian).
$Nab = Nba \implies Naba^{-1}b^{-1} = N$ (laws of coset multiplication)
$aba^{-1}b^{-1} \in N$.
If $H = S_5$, then $[H,H] = A_5$ ($S_5/A_5$ is abelian, so $[S_5,S_5]$ is contained in $A_5$, and every $3$-cycle is a commutator, for example:
$(a\ b\ c) = [(a\ c\ b),(b\ c)] = (a\ c\ b)(b\ c)(a\ b\ c)(b\ c) = (a\ c\ b)^2$).
Thus $N$ can only be $A_5$ or $S_5$, both of which contain every $3$-cycle.
The case $H = A_5$ is more subtle. Here, we must investigate the group $[A_5,A_5]$.
Note: $[(a\ b\ e),(a\ c\ d)] = (a\ b\ e)(a\ c\ d)(a\ e\ b)(a\ d\ c) = (a\ c\ d\ b\ e)(a\ d\ c\ e\ b) = (a\ b\ c)$.
(notice how showing this uses $5$ letters).
This shows $[A_5,A_5]$ contains every $3$-cycle, and so must be $A_5$ itself.
So for $H = A_5$, the only possibility is $N = A_5 = H$.