Isometry group of Minkowski Space from the book "Quantum Field theory and The Standard Model - Schwartz M.D":

"The group of translations and Lorentz transformations is called the Poincare group $ISO(1,3)$ (the isometry group of Minkowski space)."

I understand that Poincare group consists of translations and Lorentz transformation, but what is $ISO(1,3)$? What does isometry group of Minkowski space means? 
 A: There are two questions. First $ISO(1,3)$ as a notation for the Poincare group is confusing/misleading and should normally be avoided. The "I" is meant for "inhomogenous" because in the time of Wigner and Bargmann, the Poincare group was called the "inhomogenous Lorentz group" 
As you should know, there are three Lorentz groups, $O(1,3)$ - the full Lorentz group, this is traditionally denoted by $\mathcal L$, then $SO(1,3)$ - the Lorentz group of transformations with det =1, this is traditionally denoted by $\mathcal L_+$, then the so-called restricted Lorentz group, $SO^\uparrow (1,3)$, also denoted by $\mathcal L_{+}^{\uparrow}$. To each of the three groups, one defines their action on the $\mathbb R^4$ manifold of the abelian translations group and from here forms 3 semidirect products.
In physics, the isometry group of Minkowski spacetime is obviously the full Poincare group $\mathcal P = \mathbb R^4 \rtimes \mathcal L$, but this is too big for our purposes (Standard Model) because this contains the discrete PT transformations which can be individually broken. Therefore, the isometry group whose representations define the fundamental particles is only the subgroup $\mathcal P_{+}^{\uparrow} = \mathbb R \rtimes \mathcal L_{+}^{\uparrow}$. 
The second question: read the Wiki on the "isometry group". This is typically the group of transformations which leave invariant the norm of vectors and the distance between points. Actually, the definition itself of the Poincare group starts from this part. If you know a little bit of differential geometry, you can read these notes: http://www.physics.usu.edu/Wheeler/GenRel2013/Notes/GRKilling.pdf
A: $$ISO(3,1)~\cong~ O(3,1) \ltimes \mathbb{R}^{3,1}$$ is not named after the word isometry, although it is the isometry group of Minkowski spacetime $\mathbb{R}^{3,1}$. Rather it stands for inhomogeneous $O(3,1)$ or inhomogeneous Lorentz group, i.e. the group of inhomogeneous Lorentz transformations, aka. the Poincare group. (It would have been more logical to name it $IO(3,1)$, but unfortunately this is not recognized.) It is the group of spacetime transformations that respect the Minkowski metric.
A: A more direct (and transparent) answer is this. SO(3,1) is the group that leaves the quadratic form
$$dx^2 + dy^2 + dz^2 ‒ dw^2$$
invariant. The most general infinitesimal linear transform is given by:
$$Δdx = ω_y dz ‒ ω_z dy + ψ_x dw,$$
$$Δdy = ω_z dx ‒ ω_x dz + ψ_y dw,$$
$$Δdz = ω_x dy ‒ ω_y dx + ψ_z dw,$$
$$Δdw = ψ_x dx + ψ_y dy + ψ_z dz,$$
and has 6 parameters comprising the vectors $ = (ω_x,ω_y,ω_z),  = (ψ_x,ψ_y,ψ_z)$.
In physical applications, where $ = (x,y,z)$ is the position vector and $w = ct$, with $c$ being light speed and $t$ the time, and rewriting $$ as the vector $ = (υ_x,υ_y,υ_z) ≡ ‒c$, and setting $α = 1/c^2$ these transforms may be written
$$Δd = ×d ‒  dt,$$
$$Δdt = ‒α·d.$$
In that case, the transforms are equivalently chracterized as those which preserve the ‟proper time” invariant:
$$ds^2 ≡ dt^2 ‒ α\left(dx^2 + dy^2 + dz^2\right),$$
and the parameters are interpreted as $$ for infinitesimal Spatial Rotations and $$ for infinitesimal Boosts.
Since the transforms involve coordinate differentials, then they may be integrated to yield the transforms for the coordinates themselves:
$$Δ = × ‒ t + ,$$
$$Δt = ‒α· + τ.$$
The resulting constants of integration are then interpreted as $ = (ε_x, ε_y, ε_z)$ for infinitesimal Spatial Translations, and $τ$ for infinitesimal Time Translations.
Defining $ε_w = cτ$, and restoring $w$, the original transforms can be written:
$$Δx = ω_y z ‒ ω_z y + ψ_x w + ε_x,$$
$$Δy = ω_z x ‒ ω_x z + ψ_y w + ε_y,$$
$$Δz = ω_x y ‒ ω_y x + ψ_z w + ε_z,$$
$$Δw = ψ_x x + ψ_y y + ψ_z z + ε_w,$$
the result now having 10 parameters, comprising the 6-vector $\bar{ω} ≡ (, )$ and the 4-vector $\bar{ε} ≡ (ε_x,ε_y,ε_z,ε_w)$. This is the inhomogeneous form of SO(3,1) ― or ISO(3,1).
One can derive the finite forms of the respective infinitesimal transforms by Taylor expanding the transform as
$$(,t) → (1 + λΔ + \frac{(λΔ)^2}{2!} + \frac{(λΔ)³}{3!} + ⋯)(,t).$$
I'll describe this in more detail here, in terms of the physical units, appealing to the physical interpretation.
For translations, one has
$$Δ(,t) = (,τ),$$
$$Δ^2(,t) = Δ(,τ) = (,0),$$
which results in $Δ^n(,t) = (,0)$, for $n = 2,3,4,⋯$. Thus
$$(,t) → ( + λ, t + λτ) = ( + , t + b),$$
where $(,b) = (λ,λτ)$.
For rotations, one has
$$Δt = 0,$$
$$Δ = ×,$$
$$Δ^2 = ×Δ = ×(×),$$
$$Δ^3 = ×(×) = ×(×(×)) = ‒||^2× = ‒||^2Δ,$$
resulting in $Δ^{2+n} = ‒||^2Δ^n$, for $n = 1,2,3,⋯$. Thus
$$(,t) → ( + S× + E×(×),t),$$
where we define the functions
$$e_0(χ,λ) = 1 + χ \frac{λ^2}{2!} + χ^2 \frac{λ^4}{4!} + ⋯$$
$$e_1(χ,λ) = λ + χ \frac{λ^3}{3!} + χ^2 \frac{λ^5}{5!} + ⋯$$
$$e_2(χ,λ) = \frac{λ^2}{2!} + χ \frac{λ^4}{4!} + χ^2 \frac{λ^6}{6!} + ⋯$$
define $(S,E) = \left(e_1\left(-||^2,λ\right), e_2\left(-||^2,λ\right)\right)$ and where we also note the identities
$$e_0 ‒ χe_2 = 1,$$
$${e_1}^2 = e_2(e_0 + 1) ⇒ (e_0)^2 ‒ χ{e_1}^2 = 1.$$
For $χ = ‒||^2$, $(S,E)$ resolve into trigonometric functions
$$(S,E) = \left(\frac{\sin (λ||)}{||}, \frac{1 ‒ \cos (λ||)}{||^2}\right).$$
So, defining the angle $θ = λ||$, and the angle vector $ = λ$, we may write the finite form of the transform as:
$$(,t) → \left( + \frac{\sin θ}{θ} × + \frac{1 ‒ \cos θ}{θ^2} ×(×),t\right).$$
For boosts, we have
$$Δ(,t) = (‒t, ‒α·),$$
$$Δ^2(,t) = (‒Δt, ‒α·Δ) = \left(α·, α||^2t\right) = \left(α·, α||^2 t\right),$$
$$Δ^3 = α·Δ = ‒α||^2t = α||^2 Δ,$$
resulting in $Δ^{n+2}(,t) = α||^2 Δⁿ(,t)$, for $n = 1,2,3,⋯$, and can write
$$(,t) → ( ‒ St + Eα·, Ct ‒ Sα·),$$
where now with $χ = α||^2$
$$(C,S,E) = (e_0(χ,λ), e_1(χ,λ), e_2(χ,λ)),$$
which reduce to hyperbolic functions
$$(C,S,E) = \left(\cosh (\sqrt{α}u), \frac{\sinh (\sqrt{α}u)}{\sqrt{α}||}, \frac{\cosh (\sqrt{α}u) ‒ 1}{α||^2}\right).$$
The parameter $u ≡ ||λ$ is called the rapidity of the boost. The speed associated with the boost is given by
$$v ≡ \frac{\tanh (\sqrt{α}u)}{\sqrt{α}} = c \tanh {\frac{u}{c}}$$
and the corresponding velocity is given by the vector $ ≡ v/||$, so that one can express the transform as (using the identity $χE = C ‒ 1$):
$$(,t) → \left( ‒ Ct + (C ‒ 1)\frac{·}{v^2}, C(t ‒ α·)\right),$$
where
$$C = \cosh (\sqrt{α}u) = \frac{1}{\sqrt{1 ‒ \tanh^2 (\sqrt{α}u)}} = \frac{1}{\sqrt{1 ‒ αv^2}} = \frac{1}{\sqrt{1 ‒ v^2/c^2}},$$
and $S = C$.
That's the boost associated with the Lorentz transform. It is usually resolved, for $$, in terms of components, setting the $x$ component parallel to the boost direction $$ and the $y$ and $z$ components perpendicular to it. In that case, the transforms reduce to:
$$t → C(t ‒ αvx) = \frac{t ‒ vx/c^2}{\sqrt{1 ‒ v^2/c^2}},$$
$$x → x ‒ Cvt + (C ‒ 1)x = C(x ‒ vt) = \frac{x ‒ vt}{\sqrt{1 ‒ v^2/c^2}},$$
$$(y,z) → (y,z).$$
The transforms generated by Taylor expansions from the identity transform comprise the ‟connected” part of the groups SO(3,1) and ISO(3,1) respectively. As Lorentz, himself, noted in his writings c. 1900, one also has the following two transforms which preserve the invariant:
$$T: (,t) → (,‒t),$$
$$P: (,t) → (‒,t),$$
respectively for Time Reversal and Space Reversal.
So a general element of SO(3,1) and ISO(3,1) is of the form $Λ T^{n_0} P^{n_1}$, consisting of a transform $Λ$ from the connected part of the respective group, and powers $(n_0,n_1) ∈ {0,1}$ respectively of the discrete transforms $T$ and $P$.
