For $q \in \mathbb{N}$ let $\Gamma(q)=\operatorname{ker}\left(\mathrm{SL}_2(\mathbb{Z}) \rightarrow \mathrm{SL}_2(\mathbb{Z} / q \mathbb{Z})\right)$
Prove that the Alternating group $A_n$, for $n \geq 7$ is not isomorphic to any composition factor in a composition series of $\mathrm{SL}_2\left(\mathbb{Z} / p^t \mathbb{Z}\right)$.
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Solution: To prove that the alternating group $A_n$, for $n \geq 7$, is not isomorphic to any composition factor in a composition series of $\mathrm{SL}_2\left(\mathbb{Z} / p^t \mathbb{Z}\right)$, we will first show that $A_n$ is simple for $n \geq 7$. Then, we will analyze the possible composition factors of $\mathrm{SL}_2\left(\mathbb{Z} / p^t \mathbb{Z}\right)$ and show that none of them can be isomorphic to $A_n$.
- $A_n$ is simple for $n \geq 7$.
This is a well-known result in group theory. The proof is not too difficult, but it is somewhat involved. The main idea is to show that any nontrivial normal subgroup $N$ of $A_n$ must contain a 3-cycle, and since the 3-cycles generate $A_n$, it follows that $N = A_n$. Thus, $A_n$ is simple.
- Analyzing the composition factors of $\mathrm{SL}_2\left(\mathbb{Z} / p^t \mathbb{Z}\right)$.
Let $G = \mathrm{SL}_2\left(\mathbb{Z} / p^t \mathbb{Z}\right)$. We know that $G$ is a finite group and its order is given by $|G| = p^t(p^t - 1)(p^{t-1} + 1)$. Now, let's consider a composition series of $G$:
$$\{1\} = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G$$
The composition factors are the quotient groups $G_i / G_{i-1}$, and by Jordan-Hölder theorem, they are unique up to isomorphism. We will now analyze the possible composition factors of $G$.
(a) If $p$ is odd, then the group $G$ has a unique (up to isomorphism) non-abelian simple group of order $p(p^2 - 1)$ as a composition factor, which is isomorphic to $\mathrm{PSL}_2(\mathbb{F}_p)$. This is because the derived subgroup of $G$ is $\mathrm{SL}_2(\mathbb{F}_p)$, and its quotient by the center is $\mathrm{PSL}_2(\mathbb{F}_p)$. Note that $\mathrm{PSL}_2(\mathbb{F}_p)$ has order $p(p^2 - 1)/2$, which is not divisible by $n!$ for $n \geq 7$. Therefore, $A_n$ cannot be isomorphic to this composition factor.
(b) If $p = 2$, then the group $G$ has a unique (up to isomorphism) non-abelian simple group of order $2^t(2^t - 1)$ as a composition factor, which is isomorphic to $\mathrm{PSL}_2(\mathbb{F}_{2^t})$. This is because the derived subgroup of $G$ is $\mathrm{SL}_2(\mathbb{F}_{2^t})$, and its quotient by the center is $\mathrm{PSL}_2(\mathbb{F}_{2^t})$. Note that $\mathrm{PSL}_2(\mathbb{F}_{2^t})$ has order $2^t(2^t - 1)/2$, which is not divisible by $n!$ for $n \geq 7$. Therefore, $A_n$ cannot be isomorphic to this composition factor.
The remaining composition factors of $G$ are cyclic groups of prime order. Since $A_n$ is non-abelian and simple for $n \geq 7$, it cannot be isomorphic to any of these cyclic groups.
In conclusion, the alternating group $A_n$, for $n \geq 7$, is not isomorphic to any composition factor in a composition series of $\mathrm{SL}_2\left(\mathbb{Z} / p^t \mathbb{Z}\right)$.