A set of basic abstract algebra exercises

A set of basic abstract algebra exercises

I wanted to review some basic abstract algebra. Here's a few problems for
which I am seeking solution verification. Thank you very much in advance!
$\textbf{Problem:}$ Let $H$ be a subgroup of $G$, and let $X$ denote the
set of all the left cosets of $H$ in $G$. For each element $a \in G$,
define $\rho_{a}: X \rightarrow X$ as follows: $$\rho_{a} (xH) = (ax) H$$.
Prove that $\rho_{a}$ is a permutation of $X$ for each $a \in G$.
Prove that $h: G \rightarrow S_{X}$ defined by $h(a)=\rho_{a}$ is a
homomorphism.
Prove that the set $\{a \in H : xax^{-1} \in H \> \forall x \in G\}$ is
the kernel of $h$.
$\textbf{Solution:}$
Choose any $a \in G$. We first show that $\rho_{a}$ is injective. So,
assume $\rho_{a}(xH) = \rho_{a}(x'H).$ Hence, $(ax)H = (ax')H$; we need to
show $ xH = x'H$. Let $g \in xH$. Then, $g = xh_0$ and $ag = (ax)h_0 =
(ax')h_1$ by our assumption. Multiplying $ag = (ax')h_1$ on the left by
$a^{-1}$ gives us that $g= x'h_1$. Thus, $g \in x'H$. A similar argument
gives us the reverse inclusion. To prove the surjectivity of $\rho_{a}$,
let $xH \in X$. Since $a^{-1}x \in G$, we have $\rho_{a} (a^{-1}x H) =
(aa^{-1}x)H = xH$. Indeed, $\rho_{a}$ is surjective.
First, we show $\rho_{ab} = \rho_{a} \circ \rho_{b}$. Let $xH$ be an
arbitrary element belonging to $X$. Observe that $$\rho_{a} \circ \rho_{b}
(xH) = \rho_{a}((bx)H) = (abx)H = \rho_{ab}(xH).$$ Thus,
$$h(ab)=\rho_{ab}=\rho_{a} \circ \rho_{b} = h(a)h(b),$$ and we conclude
that $h$ is a homomorphism.
Let $K$ denote the kernel of $h$. We show $\{a \in H : xax^{-1} \in H \>
\forall x \in G\} = K$. To start, let $k \in K$. Then,
$h(k)=\rho_{k}=\rho_{e}$, where $e$ is the identity element of $G$. Since
$\rho_{k}=\rho_{e}$, for each $xH \in X$ we have $(kx)H=xH.$ Hence, $kxh_0
= xh_1$ for some $h_0,h_1 \in H$ and $x^{-1}kx=h_{1}h^{-1}_{0}.$ This
implies $x^{-1}kx \in H$. For clarity, put $x_0 = x^{-1}$. So, $x^{-1}kx =
x_{0}kx^{-1}_0 \in H$. Indeed, $k \in\{a \in H : xax^{-1} \in H \> \forall
x \in G\}$. To prove the reverse inclusion, this time let $k \in \{a \in H
: xax^{-1} \in H \> \forall x \in G\}.$ Then, we must show $(kx)H=xH$. Let
$g\in (kx)H$. Suppose $g = kxh_0$ for $h_0 \in H$. Multiplying on the
right by $x^{-1}$, we obtain $x^{-1}g = x^{-1}kxh_0 = h_1$ for some $h_1
\in H$. Multiplying on the right by $x$, we indeed get $g=xh_1 \in xH$.
For the reverse inclusion, we let $g \in xH$ so that $g=xh_0$ for some
$h_0$. Then, $$kg=kxh_0$$ $$g^{-1}kg=g^{-1}kxh_{0}$$ $$
h_{1}=g^{-1}kxh_0$$ $$g=kxh_0h^{-1}_{1}.$$ The last line gives us that $g
\in (kx)H$ as desired. $\blacksquare$