Let $G$ be a Lie Group. A one-parameter subgroup of $G$ is a group homomorphism $$\phi\colon\mathbb{R}\to G$$ that is also a differentiable map at the same time. We view $\mathbb{R}$ additively and $G$ multiplicatively, so that $\phi(r+s)=\phi(r)\phi(s)$

Examples.

1. If $G=\operatorname{GL}(n,k)$ where $k=\mathbb{R}$ or $\mathbb{C}$ then any one-parameter subgroup has the form $$\phi(t)=e^{tA},$$ where $A=\frac{d\phi}{dt}(0)$ is an $n\times n$ matrix over $k$ The matrix $A$ is just a tangent vector to the Lie group $\operatorname{GL}(n,k)$ This property establishes the fact that there is a one-to-one correspondence between one-parameter subgroups and tangent vectors of $\operatorname{GL}(n,k)$ The same relationship holds for a general Lie group. The one-to-one correspondence between tangent vectors at the identity (the Lie algebra) and one-parameter subgroups is established via the exponential map instead of the matrix exponential.
2. If $G=\operatorname{O}(n,\mathbb{R})\subseteq\operatorname{GL}(n,\mathbb{R})$ the orthogonal group over $R$ then any one-parameter subgroup has the same form as in the example above, except that $A$ is skew-symmetric: $A^{\operatorname{T}}=-A$
3. If $G=\operatorname{SL}(n,\mathbb{R})\subseteq\operatorname{GL}(n,\mathbb{R})$ the special linear group over $R$ then any one-parameter subgroup has the same form as in the example above, except that $\operatorname{tr}(A)=0$ where $\operatorname{tr}$ is the trace operator.
4. If $G=\operatorname{U}(n)=\operatorname{O}(n,\mathbb{C})\subseteq\operatorname{GL}(n,\mathbb{C})$ the unitary group over $C$ then any one-parameter subgroup has the same form as in the example above, except that $A$ is skew-Hermitian: $A=-A^{*}=-\overline{A}^{\operatorname{T}}$ and $\operatorname{tr}(A)=0$