Paper 2019 Solution

Which of the following statements are true?
  1. If $G$ is a finite group, then there exists $n \in N$ such that $G$ is isomorphic to a subgroup of $GL_{n}(R)$.
  2. There exists an infinite group $G$ such that every element, other than the identity element, is of order 2.
  3. The group $GL_{2}(R)$ contains a cyclic subgroup of order 5.


Option A:
Let $G$ be a finite groupe and $n = \text{Card}(G)$.
Caley's theorem states that there exists $\phi \in \text{Hom}(G, S_n)$ which is injective, which means that $G$ is isomorphic to a subgroup of $S_n$.
We will make $S_n$ "act" on the vector space $\mathbb{R}^n$. We will define for every $\sigma \in S_n$ the matrix : $$M_{\sigma} = (\delta_{\sigma(i), j})$$
where $\delta$ is the Kronecker delta defined by : $$\delta_{a, b} = \begin{cases} 1 \text{ if } a = b \\ 0 \text{ otherwise} \end{cases}$$
These $M_{\sigma}$ are elements of $\text{GL}_n(\mathbb{R})$ because $\forall \sigma \in S_n, \ M_{\sigma} \times M_{\sigma^{-1}} = M_{\sigma^{-1}} \times M_{\sigma} = I_n$.
Let $\phi' : \sigma \mapsto M_{\sigma}$. $\phi' \in \text{Hom}(S_n, \text{GL}_n(\mathbb{R}))$ and is injective, so $S_n$ is isomorphic to a subgroup of $\text{GL}_n(\mathbb{R})$.
then, let $\theta = \phi' \circ \phi$. $\phi$ and $\phi'$ are injective morphisms, so $\theta$ is an injective morphism from $G$ to $\text{GL}_n(\mathbb{R})$. That means $G$ is isomorphic to a subgroup of $\text{GL}_n(\mathbb{R})$, which is, in our case, $\text{Im}(\theta)$.

Option B:
2Z mod 2 under addition is a Group of infinite order where each element is of order 2 other than identity element

Option C:
We need a $A \in GL_{2}(R)$ such that $A_{5} = I$. This is possible if we take rotation matrix as example
$$A = \begin{pmatrix}\cos 2\pi/n &-\sin2\pi/n \\ \sin2\pi/n & \cos 2\pi/n\end{pmatrix} \mapsto e^{\frac{2\pi i}{n}}$$

Let $p$ be a prime number. Let $n \in N$, $n \gt 1$. What is the order of a p-Sylow subgroup of $GL_{n}(F_{p})$?


For any $ A \in GL_{n}(F_{p}) \text{ we have} |A| \neq 0$ means all colums are linearly independent
Let how many ways we can make first column = $p^{n} - 1$ ( subtract 1 if all are identity element)
No of ways for two linearly independent column = $(P^{n} - 1) - (p - 1) = p^{n} - p $(Linear combination of two column can be generated by p - 1 element)
No of ways for thrid linearly independent column = $(P^{n} - 1) - (p^{2} - 1) = p^{n} - p^{2} $(Linear combination of three column can be generated by $p^{2} - 1$ element)
and so on
$$ \begin{equation} \begin{aligned} \text{Order of the group} & = (p^{n} - p)(p^{n} - p^{2}) \cdots (p^{n} - p^{n-1}) \\ & = p^{1+2+ ... +n-1} (p^{n} - 1)(p^{n-1} -1) \\ & = p^{\frac{n(n-1)}{2}} (p^{n} - 1)(p^{n-1} -1) \cdots (p^{1} - 1) \end{aligned} \end{equation} $$

Give an example of a 5-Sylow subgroup of $GL_{3}(F_{5})$ ?


$|GL_{n}(F_{p})| \neq 0$ means all colums are linearly independent
Let how many ways we can make first column = $p^{n} - 1$ ( subtract 1 if all are identity element)
no of ways for two linearly independent column = $(P^{n} - 1) - (p - 1) = p^{n} - p $(Linear combination of two column can be generated by p - 1 element)
no of ways for thrid linearly independent column = $(P^{n} - 1) - (p^{2} - 1) = p^{n} - p^{2} $(Linear combination of three column can be generated by $p^{2} - 1$ element)
and so on
Order of p-Shylow group =$ (p^{n} - p)(p^{n} - p^{2}) \cdots (p^{n} - p^{n-1}) = p^{1+2+ ... +n-1} (p^{n} - 1)(p^{n} -1) \cdots (p^{n} - 1) = p^{\frac{n(n-1)}{2}} ...$