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A Lie group $G$ is a smooth manifold that is also a group in the algebraic sense, such that the multiplication map $m: G \times G \to G$ and the inverse $i: G \to G$ are all smooth maps.
Let $g \in G$, we define the left translation $L_g$ and right translation $R_g$ as maps $G \to G$ by $$ L_g(h) = gh, \quad R_g(h) = hg $$
Let $G, H$ be Lie group, we say $\varphi: G \to H$ is a Lie group homomorphism, if it is a smooth map and also a group homomorphism.
Thm : Group homomorphisms are constant rank maps.
Proof: Let $\varphi: G \to H$ be a Lie group homomorphism. We use left translation to move all the maps on the tangent space $T_g G \to T_{\varphi(g)} H$ back to identity $T_e G \to T_e H$.
A Lie subgroup of $G$ is a subgroup of $G$ endowed with a topology and smooth structure making it into a Lie group and an immersed submanifold of $G$.
Prop 7.11 (Lee) : Let $G$ be a Lie group and $H \subset G$ a subgroup, which is also an embedded submanifold, then $H$ is a Lie subgroup.
Proof: This uses Corollary 5.30, which says, if $F: M \to N$ is a smooth map, $S \subset N$ an embedded submanifold, and image of $F$ is containedin $S$, then $F$ is a smooth map from $M$ to $S$. Of course, one can use slice chart for embedded manifold to prove this corollary directly, but take a look at theorem 5.29 is also useful. Back to this proposition, we just need to check that the multiplication $m: H \times H \to G$ and $i: H \to G$ has image contained in $H$, which is guaranteed by the subgroup condition.