# Is a conjugation of a Lie subgroup a Lie subgroup? [duplicate]

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Is a conjugation of a Lie subgroup a Lie subgroup?

1. Conjugation of a subgroup is a subgroup. Thus, a conjugation of a Lie subgroup is a subgroup.
2. 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.
3. 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?

## marked as duplicate by Adam Hughes, Pierre-Guy Plamondon, Aweygan, Shailesh, JunivenMar 2 '17 at 0:04

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 $$\mu'\colon H'\times H' \to H'$$ and $$\iota'\colon H'\to H'$$ are smooth. Notice that since $\varphi$ is a group homomorphism, we may write $$\iota' = \varphi\circ\iota\circ\varphi',$$ 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 $\mu'$?
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 $G$.
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.