Twisted S3 in A5

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:

Effect of subgroup operators
In the table below, we provide values specific to $$H$$.

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

Smaller subgroups
The values given below are specific to $$H$$.