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The universal cover $G$ of $SL_2(\mathbb{R})$ has no continuous injective homomorphism into any $GL_n(\mathbb{R})$. Whether it has a faithful representation into any $GL_n(k)$ is a different question, …

Each finite subgroup $G$ of $GL_2(\mathbb{Z})$ is a finite subgroup of $GL_2(\mathbb{R})$. Taking a positive definite form on $\mathbb{R}^2$ and averaging by the action of $G$ gives a positive defini …

For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward. Let $g$ have order $p^r$ in $G$ (if not then we are effectively working in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism …

Yes, you need to throw in $-I$. Check that the set of all matrices of the form $$\left(\begin{matrix} a&b\\\ c&d \end{matrix}\right)$$ with $b$ and $c$ even and $a\equiv d\equiv1$ (mod $4$) is a subgr …

It's well-known that for a natural number $n$ with prime factorization $n=\prod_i p_i^{r_i}$, all groups of order $n$ are abelian if and only if all $r_i\le 2$ and $\gcd(n,\Phi(n))=1$ where $\Phi(n)=\ …

Certainly $\mathrm{SL}(2,\mathbb{Z})$ contains a free group. For instance $\Gamma(2)$, the subgroup of all matrices congruent to the identity modulo $2$, is free of rank $2$. The matrices $\left(\begi …

The question is to prove that if $H$ and $K$ are subgroups of a finite Abelian group or orders $m$ and $n$ then $G$ has a subgroup of order $\mathrm{lcm}(m,n)$. Beals starts by doing the case where $ …

There are very simple examples with $H\cong\mathbb{Z}$. For instance let $G$ be the affine linear group over $\mathbb{Q}$ consisting of all maps $x\mapsto ax+b$ where $a\in\mathbb{Q}^*$ and $b\in\math …

Effectively you have a group $\Gamma$ and a monomorphism $\phi:\Gamma\to\Gamma$ which is not a surjection. Take the direct limit of the sequence $(\Gamma_n)$ where each $\Gamma_n=\Gamma$ and each map …

It is the case that each isotropic vector in $V$ has the form $u+w$ where $u\in U$ and $w\in W$ but $u$ and $w$ need not be isotropic. To see where $2q-1$ and $q-1$ come from, the quadratic form on $ …

The answer to your question is "there must be, it's just a question of doing the bookkeeping carefully". It's well-known that a subgroup of $\mathrm{PGL}(2,p)$ with order prime to $p$ is either cyclic …

Let $G$ be the group of affine linear maps over the Galois field $k=GF(16)$ of order $16$. The elements of $G$ are maps from $k$ to itself of the form $x\mapsto ax+b$ where $a\in k^*$ and $b\in G$. Th …

The group $G=\mathbb{Z}_{\ell n} \times_{(\mathbb{Z}_n,r)} \mathbb{Z}_{m n}$ is isomorphic to $\mathbb{Z}_{an} \times\mathbb{Z}_b$ where $a$ is the least common multiple of $\ell$ and $m$ and $b$ is t …

This is the proof that uses the lemma that if a finite group $G$ has a Sylow $p$-subgroup then so does each subgroup of $G$. To complete the proof of existence of Sylow $p$-subgroups, it suffices to s …

Suppose that $z=xy$ is in the centre where $x\in N$ and $y\in K$. Then for all $u\in K$, $uxy=xyu$. But $uxy=\phi(u)(x)uy$ so that $x=\phi(u)(x)$ (and $uy=yu$). As this is true for all $u\in K$ then b …