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.  $\blacksquare$


It looks good in its current state, but some of the arguments could be simplified. Do recall that
$$xH = yH \iff x^{-1} y \in H$$
So for the proof of the first one, we have
\begin{align}
\rho_\alpha (xH) = \rho_{\alpha}(x'H) &\iff (\alpha x) H = (\alpha x') H \\
&\iff (\alpha x)^{-1} (\alpha x') \in H \\
&\iff x^{-1} \alpha^{-1} \alpha x' \in H \\
&\iff x^{-1} x' \in H \\
&\iff xH = x'H
\end{align}
Proceeding similarly, the proof of 3 can be shortened.
