Brian Bi
$\DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\lcm}{lcm} \newcommand\divides{\mathbin |} \newcommand\ndivides{\mathbin \nmid} \newcommand\d{\mathrm{d}} \newcommand\p{\partial} \newcommand\C{\mathbb{C}} \newcommand\N{\mathbb{N}} \newcommand\Q{\mathbb{Q}} \newcommand\R{\mathbb{R}} \newcommand\Z{\mathbb{Z}} \newcommand\pref{(\ref{#1})} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Gal}{Gal}$
Exercise 2.5.1 Suppose $$G$$ is cyclic. Let $$x$$ be a generator of $$G$$. Let $$y \in G'$$. Then $$y = \varphi(x^i)$$ for some $$i$$ since $$\varphi$$ is surjective, and this equals $$\varphi(x)^i$$ since $$\varphi$$ is a homomorphism. This holds for arbitrary $$y$$, therefore $$\varphi(x)$$ generates $$G'$$.
Suppose $$G$$ is abelian. Let $$x, y \in G'$$. Since $$\varphi$$ is surjective, we can find $$a, b \in G$$ with $$\varphi(a) = x, \varphi(b) = y$$, whence $$xy = \varphi(a)\varphi(b) = \varphi(ab) = \varphi(ba) = \varphi(b) \varphi(a) = yx$$. This holds for arbitrary $$x, y$$, so $$G'$$ is abelian.
Exercise 2.5.2 $$K$$ and $$H$$ both contain the identity, so $$K \cap H$$ contains the identity. For all $$x \in K \cap H$$ we have $$x \in K$$ and $$x \in H$$, so $$x^{-1} \in K$$ and $$x^{-1} \in H$$, so $$x^{-1} \in K \cap H$$. Finally, for all $$x, y \in K \cap H$$, we have $$x, y \in K$$ and $$x, y \in H$$, so $$xy \in K$$ and $$xy \in H$$, so $$xy \in K \cap H$$. Therefore $$K \cap H$$ is a subgroup. In the case where $$K$$ is a normal subgroup of $$G$$, for all $$x \in K \cap H$$ and $$g \in H$$, we have that $$gxg^{-1} \in K$$ by virtue of the normality of $$K$$ and $$gxg^{-1} \in H$$ since $$x, g \in H$$, therefore $$gxg^{-1} \in K \cap H$$; so $$K \cap H$$ is a normal subgroup of $$H$$ (but not necessarily of $$G$$).
Exercise 2.5.6 Consider the elementary matrix of the first kind with $$1$$ in position $$(r, c)$$, which we will notate as $$E^{r,c}$$. Let $$M \in GL_n(\mathbb{R})$$. Then $$E^{r,c}M$$ is the matrix obtained by adding row $$c$$ of $$M$$ to row $$r$$, while $$ME^{r,c}$$ is the matrix obtained by adding column $$r$$ of $$M$$ to column $$c$$. The two results are the same if and only if each element of row $$r$$ has the same value added to it by both multiplications and each element of column $$c$$ has the same value added to it by both multiplications. For all $$i \neq c$$, the value added to $$M_{r,i}$$ by left-multiplication is $$M_{c,i}$$ and it is unaffected by right-multiplication, therefore all such $$M_{c,i}$$ must vanish in order for $$M$$ to commute with $$E^{r,c}$$. For all $$i \neq r$$, the value added to $$M_{i,c}$$ by right-multiplication is $$M_{i,r}$$ and it is unaffected by left-multiplication, therefore all such $$M_{i,r}$$ must likewise vanish. That is, all off-diagonal elements in row $$c$$ must vanish and all off-diagonal elements in column $$r$$ must vanish. The element $$M_{r,c}$$ has $$M_{c,c}$$ added to it by left-multiplication and $$M_{r,r}$$ added by right-multiplication, so those two elements must be equal. Since $$r,c$$ can be chosen arbitrarily, it follows that all off-diagonal elements of $$M$$ vanish and all on-diagonal elements are equal to each other, that is, $$M$$ is a scalar matrix. All nonzero scalar matrices are indeed central and invertible, so such matrices for the centre of $$GL_n(\mathbb{R})$$. Note that this proof works for any field.