Twisted S3 in A5

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) symmetric group:S3 and the group is (up to isomorphism) alternating group:A5 (see subgroup structure of alternating group:A5).
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Definition

Let $G$ be the alternating group:A5, i.e., the alternating group (the group of even permutations) on the set $\{ 1,2,3,4,5 \}$. $G$ has order $5!/2 = 60$.

Consider the subgroup:

$H = H_{\{1,2,3\},\{4,5 \}} = \langle (1,2,3), (1,2)(4,5) \rangle = \{ (), (1,2,3), (1,3,2), (1,2)(4,5), (2,3)(4,5), (1,3)(4,5) \}$

One way of thinking of $H$ is as follows: we first take the symmetric group on the set $\{1,2,3 \}$, which is isomorphic to symmetric group:S3, and is a subgroup of symmetric group:S5 but not of alternating group:a5. We then define a homomorphism $\alpha$ from this to the subgroup $\{ (), (4,5) \}$ of symmetric group:S5 as follows: every even permutation goes to the identity, and every odd permutation goes to $(4,5)$. We then consider the image of the group under the map $g \mapsto g\alpha(g)$. The upshot: even permutations remain as they are, and odd permutations get multiplied by $(4,5)$.

We see by construction that the map $g \mapsto g\alpha(g)$ has trivial kernel, so $H$ is isomorphic to symmetric group:S3. It is also clear that since $g$ and $\alpha(g)$ have the same sign, the product permutation is even and hence in the alternating group.

$H$ has a conjugate corresponding to every possible partition of $\{ 1,2,3,4,5 \}$ into a subset of size 3 and a subset of size 2:

Subset of size 3	Subset of size 2	Corresponding conjugate of $H$
1,2,3	4,5	$\! \{ (), (1,2,3), (1,3,2), (1,2)(4,5), (2,3)(4,5), (1,3)(4,5) \}$
1,2,4	3,5	$\! \{ (), (1,2,4), (1,4,2), (1,2)(3,5), (2,4)(3,5), (1,4)(3,5) \}$
1,2,5	3,4	$\! \{ (), (1,2,5), (1,5,2), (1,2)(3,4), (2,5)(3,4), (1,5)(3,4) \}$
1,3,4	2,5	$\! \{ (), (1,3,4), (1,4,3), (1,3)(2,5), (3,4)(2,5), (1,4)(2,5) \}$
1,3,5	2,4	$\! \{ (), (1,3,5), (1,5,3), (1,3)(2,4), (3,5)(2,4), (1,5)(2,4) \}$
1,4,5	2,3	$\! \{ (), (1,4,5), (1,5,4), (1,4)(2,3), (4,5)(2,3), (1,5)(2,3) \}$
2,3,4	1,5	$\! \{ (), (2,3,4), (2,4,3), (2,3)(1,5), (3,4)(1,5), (2,4)(1,5) \}$
2,3,5	1,4	$\! \{ (), (2,3,5), (2,5,3), (2,3)(1,4), (3,5)(1,4), (2,5)(1,4) \}$
2,4,5	1,3	$\! \{ (), (2,4,5), (2,5,4), (2,4)(1,3), (4,5)(1,3), (2,5)(1,3) \}$
3,4,5	1,2	$\! \{ (), (3,4,5), (3,5,4), (3,4)(1,2), (4,5)(1,2), (3,5)(1,2) \}$

Arithmetic functions

Function	Value	Explanation
order of whole group	60	See alternating group:A5
order of subgroup	6	See symmetric group:S3
index of the subgroup	10
size of conjugacy class	10
number of conjugacy classes in automorphism class	1

Effect of subgroup operators

In the table below, we provide values specific to $H$.

Function	Value as subgroup (descriptive)	Value as subgroup (link)	Value as group
normalizer	the subgroup itself	current page	symmetric group:S3
centralizer	trivial subgroup	--	trivial group
normal core	trivial subgroup	--	trivial group
normal closure	the whole group	--	alternating group:A5
characteristic core	trivial subgroup	--	trivial group
characteristic closure	the whole group	--	alternating group:A5

Conjugacy class-defining functions

Conjugacy class-defining function	What it means in general	Why it takes this value
Sylow normalizer for the prime $p = 3$	Normalizer of a $p$-Sylow subgroup. Since Sylow implies order-conjugate (any two $p$-Sylow subgroups are conjugate), any two normalizers thereof are also conjugate.	A 3-Sylow subgroup is A3 in A5, given by $\{ (), (1,2,3), (1,3,2) \}$, and its normalizer is precisely this subgroup.

Related subgroups

Intermediate subgroups

There are no intermediate subgroups, because this subgroup is a maximal subgroup.

Smaller subgroups

The values given below are specific to $H$.

Value of smaller subgroup (descriptive)	Isomorphism class of smaller subgroup	Number of conjugacy classes of smaller subgroup fixing subgroup and whole group	Smaller subgroup in subgroup	Smaller subgroup in whole group
$\{ (), (1,2,3), (1,3,2) \}$	cyclic group:Z3	1	A3 in S3	A3 in A5
$\{ (), (1,2)(4,5) \}$	cyclic group:Z2	3	S2 in S3	subgroup generated by double transposition in A5

Description in terms of alternative interpretations of the whole group

Description of $G$	Corresponding description of $H$
special linear group of degree two over field:F4	Group of monomial matrices of determinant 1, i.e., matrices with determinant 1 and where each row has exactly one nonzero entry and each column has exactly one nonzero entry. Alternatively, one of the conjugate subgroups can be thought of as a copy of $SL(2,2)$ in $SL(2,4)$, via the embedding of field:F2 in field:F4.
projective special linear group of degree two over field:F5	If we think of the field $\mathbb{F}_{25}$ as a two-dimensional vector space over field:F5, its multiplicative group embeds in $GL(2,5)$. The intersection with $SL(2,5)$ corresponds to the elements whose sixth power is the identity. Projecting down to $PSL(2,5)$ gives a cyclic group of order three, which is precisely the subgroup A3 in A5. There is an automorphism arising from the Frobenius upstairs, and introducing this as an element of $PSL(2,5)$ gives the whole group.

Subgroup properties

Normality-related properties

Property	Meaning	Satisfied?	Explanation	Comment
normal subgroup	equals all its conjugate subgroups	No	(see above for other conjugate subgroups)
2-subnormal subgroup	normal subgroup in its normal closure	No
subnormal subgroup	series from subgroup to whole group, each normal in next	No
contranormal subgroup	normal closure is whole group	Yes
abnormal subgroup	For any element $g \in G$, we have $g \in \langle H, gHg^{-1} \rangle$	Yes	Follows from being a Sylow normalizer, see Sylow normalizer implies abnormal
weakly abnormal subgroup		Yes	Follows from being abnormal
self-normalizing subgroup	equals normalizer in the whole group	Yes
self-centralizing subgroup	contains its centralizer in the whole group	Yes
subgroup whose join with any distinct conjugate is the whole group	join of the subgroup with any distinct conjugate subgroup is the whole group	Yes
maximal subgroup	no proper subgroup containing it	Yes		Note that in a finite solvable group, any maximal subgroup has prime power index (see maximal subgroup has prime power index in finite solvable group). The fact that we have a maximal subgroup here whose index is not a prime power is consistent with the fact that alternating group:A5 is not solvable.
pronormal subgroup	any conjugate to it is conjugate in their join	Yes
weakly pronormal subgroup		Yes
paranormal subgroup		Yes
polynormal subgroup		Yes

Resemblance-based properties

Property	Meaning	Satisfied?	Explanation	Comment
order-conjugate subgroup	conjugate to any subgroup of the whole group of the same order	Yes	See subgroup structure of alternating group:A5
isomorph-conjugate subgroup	conjugate to every subgroup of the whole group isomorphic to it	Yes	See subgroup structure of alternating group:A5
automorph-conjugate subgroup	conjugate to every subgroup of the whole group automorphic to it	Yes	Follows from being a Sylow normalizer	Sylow implies automorph-conjugate