I am currently trying to prove that the two following commutator relationships are equivalent (for an operator $\hat{A}(s)$ that depends on a continuous parameter $s$), so if one holds the other one should hold as well:
$$0=\left[\frac{d\hat{A}(s)}{ds},\exp(\hat{A}(s))\right]$$
and
$$0=\left[\frac{d\hat{A}(s)}{ds},\hat{A}(s)\right].$$

**Proof from the second relation to the first**

That the second equation implies the first one is easy to see. Just start from the first equation and fill in the series expansion of the exponential, yielding:
$$0=\sum\limits\_{n=0}^\infty\frac{1}{n!}\left[\frac{d\hat{A}(s)}{ds},\hat{A}^n(s)\right].$$ The above commutator can be expanded in terms of the second commutator, proving that if the second relation holds, the first one is implied.

**Proof from the first equation to the second**

The other way around is much harder, and it's here that I am stuck! I could re-enter the series and expand the first relation as:
$$0=\left[\frac{d\hat{A}(s)}{ds},\exp(\hat{A}(s))\right]=\left[\frac{d\hat{A}(s)}{ds},\hat{1}\right]+\left[\frac{d\hat{A}(s)}{ds},\hat{A}(s)\right]+\frac{1}{2}\left[\frac{d\hat{A}(s)}{ds},\hat{A}(s)^2\right]+...,$$
but I don't see how this might help me.

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> So my question is: do these relations truly imply each other? And if not, what are the conditions such that they do ? So, if one of the two is true, does this automatically imply the other?
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