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Junkie's comment answers both parts of the first question, since $\pm \Gamma_{0}(3)$ contains no element of order $4$ (its image after reduction (mod 3) would still have order $4$). They also provide a suggestion to deal with other primes. For other $p> 3$, I think you can do something like this. The matrices congruent to the identity (entrywise) (mod p) form a torsion free normal subgroup $N$ of ${\rm GL}(2,\mathbb{Z})$. The image of the congruence subgroup (mod $N$) is solvable, and has a normal Sylow $p$-subgroup with Abelian factor group. However, if $X$ is a subgroup of the congruence subgroup isomorphic to ${\rm GL}(2,\mathbb{Z})$, then $X/X \cap N$ contains a dihedral subgroup of order $8$.

Junkie's comment answers both parts of the first question, since $\pm \Gamma_{0}(3)$ contains no element of order $4$ (its image after reduction (mod 3) would still have order $4$). They also provide a suggestion to deal with other primes. For other $p> 3$, I think you can do something like this. The matrices congruent to the identity (entrywise) (mod p) form a torsion free normal subgroup $H$ of ${\rm GL}(2,\mathbb{Z})$. The image of the congruence subgroup (mod $H$) is solvable, and has a normal Sylow $p$-subgroup with Abelian factor group. However, if $X$ is a subgroup of the congruence subgroup isomorphic to ${\rm GL}(2,\mathbb{Z})$, then $X/X \cap H$ contains a dihedral subgroup of order $8$.

Junkie's comment answers both parts of the first question, since $\pm \Gamma_{0}(3)$ contains no element of order $4$ (its image after reduction (mod 3) would still have order $4$). They also provide a suggestion to deal with other primes. For other $p> 3$, I think you can do something like this. The matrices congruent to the identity (entrywise) (mod p) form a torsion free normal subgroup $N$ of ${\rm GL}(2,\mathbb{Z})$. The image of the congruence subgroup (mod $N$) is solvable, and has a normal Sylow $p$-subgroup with Abelian factor group. However, if $X$ is a subgroup of the congruence subgroup isomorphic to ${\rm GL}(2,\mathbb{Z})$, then $X/X \cap N$ contains a quaternion subgroup of order $8$.

Junkie's comment answers both parts of the first question, since $\pm \Gamma_{0}(3)$ contains no element of order $4$ (its image after reduction (mod 3) would still have order $4$). They also provide a suggestion to deal with other primes. For other $p> 3$, I think you can do something like this. The matrices congruent to the identity (entrywise) (mod p) form a torsion free normal subgroup $N$ of ${\rm GL}(2,\mathbb{Z})$. The image of the congruence subgroup (mod $N$) is solvable, and has a normal Sylow $p$-subgroup with Abelian factor group. However, if $X$ is a subgroup of the congruence subgroup isomorphic to ${\rm GL}(2,\mathbb{Z})$, then $X/X \cap N$ contains a dihedral subgroup of order $8$.

Junkie's comment answers both parts of the first question, since $\pm \Gamma_{0}(3)$ contains no element of order $4$ (its image after reduction (mod 3) would still have order $4$). They also provide a means to deal with other primes. For other $p> 3$, I think you can do something like this. The matrices congruent to the identity (entrywise) (mod p) form a torsion free normal subgroup $N$ of ${\rm GL}(2,\mathbb{Z})$. The image of the congruence subgroup (mod $N$) is solvable, and has a normal Sylow $p$-subgroup with a Klein 4-group as a factor group. However, if $X$ is a subgroup of the congruence subgroup isomorphic to ${\rm GL}(2,\mathbb{Z})$, then $X/X \cap N$ contains an element of order $4$.

Junkie's comment answers both parts of the first question, since $\pm \Gamma_{0}(3)$ contains no element of order $4$ (its image after reduction (mod 3) would still have order $4$). They also provide a suggestion to deal with other primes. For other $p> 3$, I think you can do something like this. The matrices congruent to the identity (entrywise) (mod p) form a torsion free normal subgroup $N$ of ${\rm GL}(2,\mathbb{Z})$. The image of the congruence subgroup (mod $N$) is solvable, and has a normal Sylow $p$-subgroup with Abelian factor group. However, if $X$ is a subgroup of the congruence subgroup isomorphic to ${\rm GL}(2,\mathbb{Z})$, then $X/X \cap N$ contains a quaternion subgroup of order $8$.