Some further information to add to qm-arv's correct answer, some that may help you now, some later on as you get more of a grasp on the subject matter.
A Lie group is essentially a group that is also a manifold, i.e. it can be labelled patchwise by co-ordinates and we also have the further requirement that the group operations (multiplication and inverse) induce continuous functions of these co-ordinates. That is, we can write expressions for the co-ordinates of a product as a function of the co-ordinates of the multiplicands and the expression we get defines a continuous function; likewise for the inverse.
Where this is relevant to the Schrödinger equation is that, for a time-invariant Hamiltonian, the Schrödinger equation describes the evolution of the quantum state of a system when the system is undisturbed; therefore the operator $U(\tau)$ describing evolution through some time interval $\tau$ must be the same whatever the beginning time for the evolution. So if we think about a time interval of length $\tau+s$, it is equivalent to the evolution for time interval $\tau$, followed by the evolution for time interval $s$ (or contrariwise) and we arrive at the conclusion that:
$$U(\tau+s) = U(\tau)\,U(s)\tag{1}$$
That is, the set $\{U(\tau):\,\tau\in\mathbb{R}\}$ of all evolution operators for this undisturbed system for all time intervals must form a group, called, naturally enough, a one parameter group. This is always an Abelian (commutative) group.
We're used to Nature's being continuous in her actions, so it's natural to postulate that $U(\tau)$ is a continuous function of the evolution duration $\tau$.
Now if our quantum system is a finite dimensional one, all the operators in (1) have commuting, finite dimensional matrices. We can manipulate square matrices exactly like scalars, so we can quickly deduce, as one can for real numbers, that the only continuous function that fulfils (1) is a function of the form
$$U(\tau) = \exp(K\,\tau) = \mathrm{id} + K\,\tau + K^2 \frac{\tau^2}{2!}+\cdots;\;\forall \tau\in\mathbb{R}\tag{2}$$
for some constant square matrix $K$. The series is universally convergent for finite dimensional square matrices. Also, for $\tau$ small enough, so that $U(\tau)$ must approach the identity matrix, the matrix logarithm is uniquely defined:
$$\tau\, K = U(\tau)-\mathrm{id} - \frac{(U(\tau)-\mathrm{id})^2}{2} + \frac{(U(\tau)-\mathrm{id})^3}{3}-\cdots;\; \left\|U(\tau)-\mathrm{id}\right\|<1\tag{3}$$
Lie groups always have at least one such one parameter group threading through them; in fact, every member near enough to the identity must be a member of one of these one parameter groups, so another characterization of the Lie algebra $\mathfrak{h}$ in a Lie group $\mathfrak{H}$ is the set of all logarithms of Lie group members $U\in\mathfrak{H}$ such that $\|U-\mathrm{id}\|<1$ together with all scalar multiples of these logarithms. This also leads to an equivalent characterization that the Lie algebra is:
$$\mathfrak{h} = \left\{K|\;\exp(\tau\,K)\in\mathfrak{H}\;\forall\,\tau\in\mathbb{R}\right\}\tag{4}$$
For quantum state vectors, their norms must be conserved by any evolution, so the operators concerned must all be unitary. This means that our logarithms must all be skew-Hermitian, or of the form $K=i\,H$ where $H$ is Hermitian as you have shown and qm-arv's correct answer has stated. In the unitary group, which is compact and connected, every Lie group member is the exponential of a Lie algebra member; this is true for the identity-connected component of all compact Lie groups but some noncompact groups have identity-connected subgroups with members which cannot be written as an exponential of a Lie algebra member. Of course, no element $\gamma$ that is outside the identity connected component of any Lie group can be exponential $\exp(H)$ of a Lie algebra element $H$, because such an element is connected to the identity by the path $\{\exp(t\,H)|\;t\in[0,\,1]\}$.
In the infinite dimensional case, all of this still holds, thanks to the remarkable Stone's Theorem on One Parameter Unitary Groups. This remarkable theorem shows that for any Hilbert space, as long as we talk about one parameter groups where the multiplication in (1) is strongly continuous, there is a one-to-one, onto correspondence between all such groups within the Hilbert space and the (possibly unbounded) self adjoint operators on the Hilbert space. That is, we always have $U(\tau) = \exp(i\,\tau\,H)$ for some, possibly unbounded, self adjoint operator $H$. Even weaker conditions than strong continuity can make this theorem work if the Hilbert space in question is separable, as it always is in quantum mechanics.
Where we are talking about one parameter groups of time evolution operators, the corresponding self adjoint operators are the Hamiltonians defined by isolated quantum state evolutions.
As another example of the use of this powerful theorem in quantum mechanics, we have the strongly continuous group of unitary translation operators defined on the Hilbert space $L^2$ functions by $\tilde{U}(\tau)\, \psi(x) = \psi(x+\tau)$. There is, by the theorem, a self adjoint operator $D$ such that $U(\tau) = \exp(i\,\tau\,D)$.
When we restrict $D$ to smooth functions, $D$ can be written as $D = -i\,\frac{\mathrm{d}}{\mathrm{d} x}$, and we retrieve the Taylor series:
$$\psi(x+\tau) = \psi(x) + i\,\tau\,D\,\psi(x) - \frac{\tau^2}{2!} D^2\,\psi(x) +\cdots = \psi(x) + \tau\,\frac{\mathrm{d}}{\mathrm{d} x} \psi(x) + \frac{\tau^2}{2!}\,\frac{\mathrm{d}^2}{\mathrm{d} x^2} \psi(x) + \cdots\tag{6}$$
elegantly showing the correspondence between the momentum operator $-i\,\frac{\mathrm{d}}{\mathrm{d} x}$ and translation in space.
