Alternating Group on 4 Letters/Normality of Subgroups

Normality of Subgroups of the Alternating Group on $4$ Letters
Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:

The normality status of the non-trivial proper subgroups of $A_4$ is as follows:

Order $2$ subgroups:

Order $3$ subgroups:

Order $4$ subgroup:

Proof
Testing one of the left cosets of $T = \set {e, t}$ against its corresponding right coset:

The left coset does not equal the right coset and so $T$ is not normal in $A_4$.

Testing one of the left cosets of $U = \set {e, u}$ against its corresponding right coset:

The left coset does not equal the right coset and so $U$ is not normal in $A_4$.

Testing one of the left cosets of $V = \set {e, v}$ against its corresponding right coset:

The left coset does not equal the right coset and so $V$ is not normal in $A_4$.

Testing one of the left cosets of $P = \set {e, a, p}$ against its corresponding right coset:

The left coset does not equal the right coset and so $P$ is not normal in $A_4$.

Testing one of the left cosets of $Q = \set {e, c, q}$ against its corresponding right coset:

The left coset does not equal the right coset and so $Q$ is not normal in $A_4$.

Testing one of the left cosets of $R = \set {e, d, r}$ against its corresponding right coset:

The left coset does not equal the right coset and so $R$ is not normal in $A_4$.

Testing one of the left cosets of $S = \set {e, b, s}$ against its corresponding right coset:

The left coset does not equal the right coset and so $S$ is not normal in $A_4$.

The cosets of $K = \set {e, t, u, v}$ are as follows:

The left cosets equal the right cosets and so $K$ is normal in $A_4$.