Is a conjugation of a Lie subgroup a Lie subgroup? Is a conjugation of a Lie subgroup a Lie subgroup? 


*

*Conjugation of a subgroup is a subgroup. Thus, a conjugation of a Lie subgroup is a subgroup. 

*Conjugation is a diffeomorphism. Thus, a conjugation of a Lie subgroup is the image of a manifold by a diffeomorphism, and so it is an embedded submanifold.

*Finally I need to prove the smoothness of the group multiplication and inverse operation. But I'm not sure that the restrictions of them on the submanifold are still smooth. How can I prove this?

 A: Let $G$ be a Lie group, $H$ a Lie subgroup of $G$ and $g\in G$. You are asking if $gHg^{-1}$ is a Lie subgroup of $G$. The answer is yes:
The map
$$C_g:G\to G,\quad a\mapsto gag^{-1}$$
is a Lie group isomorphism (a diffeomorphism and a group isomorphism). Indeed, $C_{g^{-1}}$ provides a smooth inverse. Thus, $C_g(H)=gHg^{-1}$ is a Lie subgroup of $G$.
A: Consider a Lie subgroup $H$ of the Lie group $G$ and fix $g_0\in G$.  As you mention, the conjugation map $\varphi\colon h\mapsto g_0hg_0^{-1}$ is a diffeomorphism of $H$ onto its image $H'$.  Your question is then whether
\begin{equation}
\mu'\colon H'\times H' \to H'
\end{equation}
and
\begin{equation}
\iota'\colon H'\to H'
\end{equation}
are smooth.  Notice that since $\varphi$ is a group homomorphism, we may write
\begin{equation}
\iota' = \varphi\circ\iota\circ\varphi',
\end{equation}
where $\iota$ is the inverse map on $H$.  Since the three maps on the right are smooth, so is $\iota'$.  Can you say something similar for $\mu'$?
A: Assume $a \in G$ and that $(H,\varphi)$ is a Lie subgroup of $G$. If $c_a\colon G \to G$ is conjugation by $a$, we want to see that $(c_a(H), \varphi \circ c_a^{-1})$ is a Lie subgroup of $G$. Since $c_a$ is a diffeomorphism and $c_a(H)$ is a subgroup of $G$, we only have to see that multiplication is smooth (smoothness of inversion follows by the Inverse Function Theorem).
If $m_G$ is multiplication in $G$, etc, we have that $$m_{c_a(H)}(x,y) = m_G(\varphi \circ c_a^{-1}(x), \varphi\circ c_a^{-1}(y)) = m_H \circ (c_a^{-1} \times c_a^{-1})(x,y)$$and so $m_{c_a(H)}$ is smooth, since both $m_H\colon H \times H \to H$ and $c_a^{-1}\times c_a^{-1}\colon c_a(H) \times c_a(H) \to H \times H$ are smooth.
