Let $G$ be a Lie group ang let $\mathcal{g} = Lie(G)$ be the Lie algebra of $G$. If X and Y are left- invarient vector fields and if X' and Y' are right-invariant vector fields such that $$X_{e} = X'_{e} \quad , \quad Y_{e} = Y'_{e},$$ where $e$ is the identity element of $G$. Then $$[X , Y]_{e} = - [X' , Y']_{e}.$$ I don't know how one can prove it.
Thank you for any help.
First, I will assume that you know that given a Lie group $G$, the real vector space of all left-invariant vector fields with the Lie bracket form a Lie algebra $\mathfrak{g}$. The reason this is a Lie algebra is that a vector field is left-invariant iff it is $L_g$-related to itself $\forall g\in G$, where $L_g$ is the left translation by $g$. In exactly the same way, the real vector space of all right-invariant vector fields with the Lie bracket form a Lie algebra $\mathfrak{g}^r$.
This being said, I will make use of three lemmas:
$ \textbf{Lemma 1}:$ If $\mu:G\times G \rightarrow G$, $\iota: G \rightarrow G$ denote the multiplication and inverse maps, respectively, then $d\mu_{(e,e)}(x,y)=x+y$, $d\iota_e(x)=-x$.
$\textbf{Proof}:$ $d\mu_{(e,e)}(x,y)= d\mu_{(e,e)}(x,0)+d\mu_{(e,e)}(0,y)$. Now observe that given $\alpha:(-\epsilon, \epsilon)\rightarrow G$ with $\alpha(0)=e, \alpha'(0)=x$, the curve $\gamma:(-\epsilon, \epsilon)\rightarrow G\times G$ given by $\gamma(t)=(\alpha(t), e)$ satisfies $\gamma(0)=(e,e), \gamma'(0)=(x,0)$.
Thus, $d\mu_{(e,e)}(x,0)=\frac{d}{dt}|_{t=0} (\mu\circ \gamma)=\alpha'(0)=x$.
For the second part, observe that $\mu\circ (Id_G\times\iota)(g)=\mu(g,g^{-1})=e$, $\forall g\in G$, i.e $\mu\circ (Id_G\times\iota)$ is constant $e$. Hence, $0=\frac{d}{dt}|_{t=0} (\mu\circ (Id_G\times\iota)\circ\alpha)= d\mu_{(e,e)}\circ d(Id_G\times\iota)_e (\alpha'(0))= d\mu_{(e,e)}(x,d\iota_e(x))=x+d\iota_e(x)$.
$\textbf{Lemma 2}:$ If $Z\in \mathfrak{g}^r$, $d\iota(Z)\in \mathfrak{g}$, where $d\iota$ denotes the pushforward by $\iota$ defined by $d\iota(Z)_g=(d\iota)_{g^{-1}}(Z_{g^{-1}})$.
$\textbf{Proof}:$ $d\iota(Z)_g=(d\iota)_{g^{-1}}(Z_{g^{-1}})=(d\iota)_{g^{-1}}(Z\circ R_{g^{-1}}(e))= (d\iota)_{g^{-1}}\circ (dR_{g^{-1}})_e(Z_e)=d(\iota\circ R_{g^{-1}})_e(Z_e)=d(L_g\circ\iota)_e(Z_e)= (dL_g)_e(-Z_e)=(dL_g)_e(d\iota(Z)_e).$
$\textbf{Lemma 3}:$ $d\iota: \mathfrak{g}^r\rightarrow \mathfrak{g}$ is a Lie algebra isomorphism.
$\textbf{Proof}$: The linearity is trivial. Lemma $\textbf{1}$ proves the injectivity, which concludes the isomorphism part.
Next, take $g\in G, f\in C^{\infty}(G)$, and $X', Y'\in \mathfrak{g}^r$ as you posted. Then:
$[d\iota(X'), d\iota(X')](f)(g)=[d\iota(X'), d\iota(X')]_g(f)=d\iota(X')_g(d\iota(Y')(f))- d\iota(Y')_g(d\iota(X')(f))=(d\iota)_{g^{-1}}(X'_{g^{-1}})(d\iota(Y')(f))-(d\iota)_{g^{-1}}(Y'_{g^{-1}})(d\iota(X')(f))= X'_{g^{-1}}(d\iota(Y')(f)\circ \iota)-Y'_{g^{-1}}(d\iota(X')(f)\circ \iota).$
But $d\iota(Y')(f)\circ \iota (g)=d\iota(Y')(f) (g^{-1})=d\iota(Y')_{g^{-1}}(f)=(d\iota)_g(Y'_g)(f)=Y'_g(f\circ \iota)=Y'(f\circ \iota)(g).$
So, substituting $d\iota(Y')(f)\circ \iota=Y'(f\circ \iota)$ we get:
$[d\iota(X'), d\iota(X')](f)(g)= X'_{g^{-1}}(Y'(f\circ \iota))-Y'_{g^{-1}}(X'(f\circ \iota))=[X',Y']_{g^{-1}}(f\circ\iota)=(d\iota)_{g^{-1}}([X',Y']_{g^{-1}})(f)=d\iota([X',Y'])_g(f)= d\iota([X',Y'])(f)(g).$
Finally, with lemmas $\textbf{1}$ and $\textbf{2}$, it is clear that $d\iota(X')=-X, d\iota(Y')=-Y$. Now, with $\textbf{3}$ we have:
$d\iota([X',Y'])=[d\iota(X'),d\iota(Y')]=[-X,-Y]=[X,Y]$. So, again by $\textbf{1}$, $[X,Y]_e=-[X',Y']_e$.