Unitary groups and infinitesimal transformations - Schwingers way of deriving Lie groups In Schwinger's source theory book, he suggests if $G_a$ are the hermitian generators of the Unitary group, then we have an infinitesimal transformation is given by :
$$ G = \sum_{a=1}^n \delta\lambda_a G_a $$
Now if we subject to the infinitesimal transformation operator to arbitrary unitary transformation of the group ($\{\lambda\}$ are the group parameters) ,
$$ U^{-1}G_aU = u(\lambda)G_b $$
Alternatively,
$$  UG_aU^{-1} = G_bv(\lambda) $$
And then, these two quantities are related by $ v = (u^T)^{-1} $
Firstly, I am not able to see why the author is trying to do this kind of unitary transform on the group element, Is it possibly to get a representation? How further do we proceed with this.
 A: As I mentioned in my comment, I believe you are talking about the adjoint representation of a Lie Group $G$ with a Lie algebra $\mathfrak{g}$:
\begin{equation}
\forall x \in \mathfrak{g}, \;\; \mathrm{Ad} \:  D(g) : x \mapsto D(g) x D^{-1}(g) \in \mathfrak{g}
\end{equation}
where $D(g)$ denotes a represention of $g \in G$. One method you can see why $D(g) x D^{-1}(g) \in \mathfrak{g}$ is by considering the above transformation:
\begin{equation}
\begin{aligned}
x' & = D(g) x D^{-1}(g) \\
&= e^{i \lambda^i G_i} x e^{-i \lambda^i G_i} \\
&= (1 + i \lambda^i G_i) x (1 - i \lambda^i G_i) + O(\lambda^2) \\
&= x - x i \lambda^i G_i + i \lambda^i G_i x + O(\lambda^2)\\
&= x + i \lambda^i [G_i, x] +O(\lambda^2) \\
&= x + i \lambda^i x^j [G_i,G_j] +O(\lambda^2)\\
&= x - \lambda^i x^j f_{ij}{}^{k} G_k +O(\lambda^2)\\
&= (x^k- \lambda^i x^j f_{ij}{}^{k})G_k +O(\lambda^2)\\
&= x'^i G_i +O(\lambda^2) 
\end{aligned}
\end{equation}
where:
\begin{equation}
x'^i \equiv x^i- \lambda^j x^k f_{jk}{}^{i} 
\end{equation}
This shows that $x'$ can also be expressed in terms of the generators. In other words, the above transformation ensures that if $x$ lives in the Lie algebra formed by the generators, then $x'$ also lives in that vector space. 
Furthermore, note that from the fifth line of the above equation, we can also write:
\begin{equation}
x' = x + i \lambda^i [G_i, x] = x+ i \mathrm{ad} \; \lambda(x) = e^{i \mathrm{ad} \; \lambda} x
\end{equation}
Thus:
\begin{equation}
\mathrm{Ad} \:  D(g) (x) = e^{i \mathrm{ad} \; \lambda} x 
\end{equation}
which shows the relation between the adjoint representation of the Lie group and the adjoint representation of the Lie algebra.
A: Good Lord! Is Schwinger still worth reading? A top physicist of course, but unfortunately right over my head! You're talking about the big A Adjoint representation as in Hunter's Answer, and there are a great many more modern, rigorous and way clearer texts on this topic. The Wikipedia page is a great start. Also see Rossmann, "Lie Groups, An Introduction ...", chapter 2 or Hall, section 4.3.
Here is just a short alternative (not completion or correction, mind) to Hunter's Answer, which is altogether complete in itself.
You may find it helpful (if your brain works anything like mine, which, I admit is a slim likelihood!) to think of the Lie algebra as the linear space of tangents, at the identity, to $C^1$ paths in a Lie group through the identity. I'm saying just forget about the "manifold" for a bit; try thinking instead of the thing as a group first with $C^1$ paths linking its points. A Lie group is a highly specialised manifold (e.g. its fundamental group is Abelian, just one of a host of special things) and does not need anything like the full machinery of differential geometry to see clearly. 
If $\sigma_X:[-1,1]\to\mathfrak{G}$ is a $C^1$ path through the identity in the Lie group $\mathfrak{G}$ with $\sigma_X(0) = \mathrm{id}$ and with tangent $X\in\mathfrak{g}={\rm Lie}(\mathfrak{G})$ there, then clearly so is: 
$$\sigma_Z:[-1,1]\to\mathfrak{G};\,\sigma_Z(\tau)=U^{-1} \sigma_X(\tau) U$$ 
and it will therefore have some tangent $Z\in\mathfrak{G}$. So the conjugation $\sigma_X\mapsto U^{-1}\sigma_X U$ clearly induces some mapping $Ad(U):\mathfrak{g}\to\mathfrak{g}$ on the Lie algebra. ($C^1$ follows from the preservation of differentiability class by the group product property of any Lie group).
Now, if you let $\sigma_Y:[-1,1]\to\mathfrak{G}$ be a second $C^1$ path through the identity with tangent $Y\in\mathfrak{g}$ there, then work out from first principles what the tangent of the following $C^1$ path through the identity:
$$\sigma_Z:[-1,1]\to\mathfrak{G};\,\sigma_Z(\tau) = U^{-1}\sigma_X(\alpha\,\tau)\,\sigma_Y(\beta\tau)U = U^{-1}\sigma_X(\alpha\,\tau)\,U\,U^{-1}\sigma_Y(\beta\tau)U$$
is, where $\alpha, \beta\in \mathbb{K}$ (the field $\mathbb{K}$ is the field underlying the Lie algebra). You will find that the tangent is $\alpha {\rm Ad}(U) X + \beta{\rm Ad}(U)  Y$, so ${\rm Ad}(U)$ is a linear, homogeneous map from $\mathfrak{g}$ into itself: in loose words: it is a matrix. 
Again, from first principles considering the $C^1$ path $V^{-1}\, U^{-1} \sigma_X U\, V$ for $U,\,V\in\mathfrak{G}$, you can show ${\rm Ad}(V\, U) = {\rm Ad}(V)\,{\rm Ad}(U)$. It also follows that ${\rm Ad}(U^{-1}) = {\rm Ad}(U)^{-1}$ so that ${\rm Ad}(U)$ is invertible. Thus inverses and products of entities of the form ${\rm Ad}(U)$ for $U\in\mathfrak{G}$ is another entity of the same kind. Therefore the set:
$${\rm Ad}(\mathfrak{G})\stackrel{def}{=}\{{\rm Ad}(U):\,U\in\mathfrak{G}\}$$
is a group of invertible matrices that act on the Lie algebra $\mathfrak{g}$, and, since ${\rm Ad}(V U) = {\rm Ad}(V){\rm Ad}(U)$, ${\rm Ad}$ is a homomorphism and the pair $({\rm Ad}, \mathfrak{g})$ is therefore a representation of the original group $\mathfrak{G}$. ${\rm Ad}(\mathfrak{G}) \cong {\rm Inn}(\mathfrak{G})$, the group of inner automorphisms of $\mathfrak{G}$ and is a matrix Lie group.
As Hunter showed you, for every $X\in\mathfrak{G}$ there is another linear map definable on the Lie algebra, the so called Lie Map of ${\rm Ad}$ (i.e. its differential at the identity):
$${\rm ad}(X):\mathfrak{g}\to\mathfrak{g};\,{\rm ad}(X) Y = \left.{\rm d}_\tau {\rm Ad}(e^{
\tau\,X}) Y\right|_{\tau=0} \stackrel{def}{=}[X,\,Y]$$
This one is never invertible (${\rm ad}(X) X = 0$). With a bit of work you can prove the following interesting facts:


*

*$\ker({\rm Ad}) = \mathscr{Z}(\mathfrak{G})$, the centre of $\mathfrak{G}$,i.e. the set of all elements $\gamma\in\mathfrak{G}$ that commute with all elements of $\mathfrak{G}$;

*The Jacobi indentity: ${\rm ad}({\rm ad}(X)\, Y)={\rm ad}([X,\,Y]) = {\rm ad}(X).{\rm ad}(Y) - {\rm ad}(Y).{\rm ad} (X)$, so that little ${\rm ad}$ is a homomorphism of Lie algebras. 

*The kernel of the homomorphism ${\rm ad}$ is the centre $\mathscr{z}(\mathfrak{g})$ of $\mathfrak{g}$, the set of all $X\in \mathfrak{g}$ such that $[X,\,Y] = 0;\,\forall Y\in \mathfrak{g}$.


Point 2. is what I call a stunner: this is the REAL meaning of the Jacobi identity (you need to do a little work to show that this is indeed the Jacobi ndentity in disguise): the Jacobi identity is precisely the assertion that 
the big A ${\rm Ad}$ representation induces a homomorphism ${\rm ad}$ between the Lie algebra $\mathfrak{g}$ of a Lie group $\mathfrak{G}$ and the Lie algebra ${\rm Lie}({\rm Ad}(\mathfrak{G}))$ of ${\rm Ad}(\mathfrak{G})$.
The Lie algebra ${\rm Lie}({\rm Ad}(\mathfrak{G}))$ is the algebra ${\rm Der}(\mathfrak{g})$ of all derivations on $\mathfrak{g}$, i.e. the set of all linear maps on $\mathfrak{g}$ that fulfill the Leibnitz product rule.
