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 $A$ be the following $3 \times 3$ matrix.
\[A=\begin{bmatrix}
1 & 1 & -1 \\
0 &1 &2 \\
1 & 1 & a
\end{bmatrix}.\]
Determine the values of $a$ so that the matrix $A$ is nonsingular.

Let $S$ be the following subset of the 3-dimensional vector space $\R^3$.
\[S=\left\{ \mathbf{x}\in \R^3 \quad \middle| \quad \mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}, x_1, x_2, x_3 \in \Z \right\}, \]
where $\Z$ is the set of all integers.
Determine whether $S$ is a subspace of $\R^3$.

Let $A$ be an $m \times n$ real matrix. Then the null space $\calN(A)$ of $A$ is defined by
\[ \calN(A)=\{ \mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}_m\}.\]
That is, the null space is the set of solutions to the homogeneous system $A\mathbf{x}=\mathbf{0}_m$.

Prove that the null space $\calN(A)$ is a subspace of the vector space $\R^n$.
(Note that the null space is also called the kernel of $A$.)

Suppose that $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_r$ are linearly dependent $n$-dimensional real vectors.

For any vector $\mathbf{v}_{r+1} \in \R^n$, determine whether the vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_r, \mathbf{v}_{r+1}$ are linearly independent or linearly dependent.

Let $\mathbf{a}$ and $\mathbf{b}$ be fixed vectors in $\R^3$, and let $W$ be the subset of $\R^3$ defined by
\[W=\{\mathbf{x}\in \R^3 \mid \mathbf{a}^{\trans} \mathbf{x}=0 \text{ and } \mathbf{b}^{\trans} \mathbf{x}=0\}.\]

Prove that the subset $W$ is a subspace of $\R^3$.

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 $|X|!$.