$A_5$ is generated by all 3 cycles There are some similiar questions, but it seems that there is some different equal definition to $A_5$.
The definition I'm using for $A_5$ is that this is the kernel of the sign homomorphism- i.e all the permutations with even number of transpositions. 
I would like to get help with the equivalence of this definition , to the definition of subgroup of $S_5$ , generated by all cycles of order 3.
EDIT- THERE IS AN ANSWER ON THE LINK IN COMMENTS
A5 is the group [3,3,3]+ or [3,5]+ in Coxeter notation.
You could make it, I suppose, out of (AB)³ = (BC)³ = (CD)³ = I, and (AC)² = (BD)² = (DA)² = I, which is what the first symbol means.
The second one means (AB)³ = (BC)^5 = (AC)² = I, which is what the second definition.
A5 is a relatively primitive group, but you can indeed generate A5 from purely cycles of three.
A5 is the group [3,3,3]+ or [3,5]+ in Coxeter notation.
You could make it, I suppose, out of (AB)³ = (BC)³ = (CD)³ = I, and (AC)² = (BD)² = (DA)² = I, which is what the first symbol means.
The second one means (AB)³ = (BC)^5 = (AC)² = I, which is what the second definition.
A5 is a relatively primitive group, but you can indeed generate A5 from purely cycles of three.