A5 in S5

This article is about a particular subgroup in a group, up to equivalence of subgroups (i.e., an isomorphism of groups that induces the corresponding isomorphism of subgroups). The subgroup is (up to isomorphism) alternating group:A5 and the group is (up to isomorphism) symmetric group:S5 (see subgroup structure of symmetric group:S5).
The subgroup is a normal subgroup and the quotient group is isomorphic to cyclic group:Z2.
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Definition

We define $G$ as symmetric group:S5 -- for concreteness, the symmetric group on the set $\{1,2,3,4,5\}$.

$H$ is alternating group:A5 -- the subgroup of $G$ comprising even permutations. A criterion for a permutation to be even, based on cycle decomposition, is that the number of cycles of even length should be even.

The quotient group is cyclic group:Z2.

Arithmetic functions

Function	Value	Explanation
order of group	120	Order is $5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120$
order of subgroup	60	Order is $5!/2=120/2=60$
index of subgroup	2	For degree $n\geq 2$, the alternating group of degree $n$ has index two in the symmetric group of degree $n$, since it has two cosets: itself (the even permutations) and the odd permutations.

Subgroup-defining functions

Subgroup-defining function	Meaning in general	Why it takes this value	GAP verification (set G := SymmetricGroup(5); H := AlternatingGroup(5);)
derived subgroup (also called commutator subgroup)	subgroup generated by commutators	Since the quotient is abelian, it contains the derived subgroup. Further, $H$ is simple non-abelian, so the derived subgroup cannot be smaller.	H = DerivedSubgroup(G); using DerivedSubgroup
hypocenter	stable member of lower central series (transfinite if necessary)	$H$ is the derived subgroup, and $[G,H]=H$
perfect core	stable member of derived series (transfinite if necessary); alternatively, join of all perfect subgroups	$H$ is perfect, and the only bigger subgroup is $G$, which is not perfect because $[G,G]=H$
Jacobson radical	intersection of all maximal normal subgroups	$H$ is the unique maximal normal subgroup ($G$ is a one-headed group). This can also be seen from the fact that it is simple, so no other maximal normal subgroup can intersect it nontrivially, and it has no normal complement, so no maximal normal subgroup can intersect it nontrivially.
socle	join of all minimal normal subgroups	It is the unique minimal normal subgroup ($G$ is a monolithic group).	H = Socle(G); using Socle
3-Sylow closure	normal closure of any 3-Sylow subgroup	Any 3-Sylow subgroup lies within $H$. Since $H$ is a minimal normal subgroup, it must be the normal closure of that subgroup.
5-Sylow closure	normal closure of any 5-Sylow subgroup	Any 5-Sylow subgroup lies within $H$. Since $H$ is a minimal normal subgroup, it must be the normal closure of that subgroup.

GAP implementation

We can define the group and subgroup naturally, using the SymmetricGroup and AlternatingGroup functions:

G := SymmetricGroup(5); H := AlternatingGroup(5);