I am getting myself confused regarding the differences between the infinitesimal generators of Lie group and the elements of the Lie algebra, likely due to the fact that I am studying from a physics perspective where authors frequently use the group and the algebra interchangeably.
Consider a specific example from QFT, the Lorentz group $SO(3,1)$. Let $\Lambda^{\mu}_{\ \nu}$ be a Lorentz transformation that is infinitesimally close to the identity. i.e. $ \Lambda^{\mu}_{\ \nu} = \delta^{\mu}_{\ \nu} + \epsilon^{\mu}_{\ \nu} $ with $\epsilon^{\mu}_{\ \nu}$ "small".
From the definition of the Lorentz group, we can show that $\epsilon^{\mu}_{\ \nu}$ is antisymmetric, the usual defining property of the Lorentz algebra. This leads me to believe that $\epsilon^{\mu}_{\ \nu}$ is an elements of the lie algebra $\mathfrak{so}(31)$. However, most authors then go on to write a general Lorentz transformation as $$ \Lambda = \text{exp}\left(i\epsilon_{\mu \nu} M^{\mu \nu} \right) $$ with $$ [M^{\mu \nu}, M^{\rho \sigma}] = i(g^{\nu \rho} M^{\mu \sigma} - g^{\mu \rho} M^{\nu \sigma} - g^{\nu \sigma} M^{\mu \rho} + g^{\mu \sigma} M^{\nu \rho}) $$ which is usually how the algebra $\mathfrak{so}(3,1)$ is defined. This leads me to believe that the elements $M^{\mu \nu}$ are the elements of the lie algebra and the $\epsilon^{\mu \nu}$ are simply the parameters that define the transformation (i.e. the angle $\theta^{12}$ of the rotation generated by $M^{12} \equiv J^3$)
The use of these two definitions confuses me, especially because in the latter example, the tensor $\epsilon^{\mu \nu}$ is still referred to as antisymmetric. If anyone could provide some clarification is would be much appreciated!
I think that I answered my own question, with the help of the commenters. My confusion stemmed from the fact that I didn't see why $\epsilon^{\mu \nu}$ had to be antisymmetric, and mistook the antisymmetry to mean that $\epsilon^{\mu \nu}$ must be an element of the lie algebra. In fact, $\epsilon^{\mu \nu}$ doesn't have to be antisymmetric, but since it is being contracted with $M^{\mu \nu}$, which is antisymmetric, any symmetric part of the tensor will vanish. Thus, we might as well take $\epsilon^{\mu \nu}$ to be antisymmetric by definition.