Let $\R^{\times}=\R\setminus \{0\}$ be the multiplicative group of real numbers.
Let $\C^{\times}=\C\setminus \{0\}$ be the multiplicative group of complex numbers.
Then show that $\R^{\times}$ and $\C^{\times}$ are not isomorphic as groups.

Let $G$ be a group and $H$ and $K$ be subgroups of $G$.
For $h \in H$, and $k \in K$, we define the commutator $[h, k]:=hkh^{-1}k^{-1}$.
Let $[H,K]$ be a subgroup of $G$ generated by all such commutators.

Show that if $H$ and $K$ are normal subgroups of $G$, then the subgroup $[H, K]$ is normal in $G$.

Let $G$ be a nilpotent group and let $H$ be a subgroup such that $H$ is a subgroup in the center $Z(G)$ of $G$.
Suppose that the quotient $G/H$ is nilpotent.

Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$.
Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.

Let $G$ be a finite group and $P$ be a nontrivial Sylow subgroup of $G$.
Let $H$ be a subgroup of $G$ containing the normalizer $N_G(P)$ of $P$ in $G$.

Let $G$ and $G’$ be groups and let $f:G \to G’$ be a group homomorphism.
If $H’$ is a normal subgroup of the group $G’$, then show that $H=f^{-1}(H’)$ is a normal subgroup of the group $G$.

Let $A$, $B$ be groups. Let $\phi:B \to \Aut(A)$ be a group homomorphism.
The semidirect product $A \rtimes_{\phi} B$ with respect to $\phi$ is a group whose underlying set is $A \times B$ with group operation
\[(a_1, b_1)\cdot (a_2, b_2)=(a_1\phi(b_1)(a_2), b_1b_2),\]
where $a_i \in A, b_i \in B$ for $i=1, 2$.

Let $f: A \to A’$ and $g:B \to B’$ be group isomorphisms. Define $\phi’: B’\to \Aut(A’)$ by sending $b’ \in B’$ to $f\circ \phi(g^{-1}(b’))\circ f^{-1}$.

\[\require{AMScd}
\begin{CD}
B @>{\phi}>> \Aut(A)\\
@A{g^{-1}}AA @VV{\sigma_f}V \\
B’ @>{\phi’}>> \Aut(A’)
\end{CD}\]
Here $\sigma_f:\Aut(A) \to \Aut(A’)$ is defined by $ \alpha \in \Aut(A) \mapsto f\alpha f^{-1}\in \Aut(A’)$.
Then show that
\[A \rtimes_{\phi} B \cong A’ \rtimes_{\phi’} B’.\]

Let $G$ be a simple group and let $X$ be a finite set.
Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$.
Then show that $G$ is a finite group and the order of $G$ divides $|X|!$.

Let $\F_p$ be the finite field of $p$ elements, where $p$ is a prime number.
Let $G_n=\GL_n(\F_p)$ be the group of $n\times n$ invertible matrices with entries in the field $\F_p$. As usual in linear algebra, we may regard the elements of $G_n$ as linear transformations on $\F_p^n$, the $n$-dimensional vector space over $\F_p$. Therefore, $G_n$ acts on $\F_p^n$.

Let $e_n \in \F_p^n$ be the vector $(1,0, \dots,0)$.
(The so-called first standard basis vector in $\F_p^n$.)

Find the size of the $G_n$-orbit of $e_n$, and show that $\Stab_{G_n}(e_n)$ has order $|G_{n-1}|\cdot p^{n-1}$.

Conclude by induction that
\[|G_n|=p^{n^2}\prod_{i=1}^{n} \left(1-\frac{1}{p^i} \right).\]