Injective Lie Group Homomorphism is Immersion?

Question

How do I prove the claim of the title?

My definition of Lie Group homomorphism is a smooth group homomorphism. I have to show that $Df(e)(v)=0$ will imply $v=0$. Viewing this as a derivation, this means $Df(e)(v)(g)=0$ for all smooth functions $g$ on $H$.

Here $f$ is the Lie group homomorphism from $G$ to $H$.

Answer 1

Suppose that $Df_e(v)=0$. Then $(\forall t\in\mathbb R):Df_e(tv)=tDf_e(v)=0$. And therefore$$(\forall t\in\mathbb R):\exp\bigl(Df_e(tv)\bigr)=e.$$But then$$(\forall t\in\mathbb R):f(e^{tv})=e$$and therefore $f$ is not injective.

Answer 2

A different approach to prove this (without the exponential map) is to use the Global Rank Theorem (Lee's book, Smooth manifolds):

Theorem 4.14 (Global Rank Theorem). Let $M$ and $N$ be smooth manifolds, and suppose $F:M\to N$ is a smooth map of constant rank.

(a) if $F$ is surjective, then it is a smooth submersion.

(b) if $F$ is injective, then it is a smooth immersion.

(c) if $F$ is bijective, then it is a diffeomorphism.

Combined with the fact that a Lie group homomorphism $\varphi:G\to H$ has constant rank. This fact follows from the equality $\varphi\circ L_g=L_{\varphi(g)}\circ\varphi$ where $g\in G$ and $L_g(h)=gh\;\forall\, h\in G$, and taking the derivative.