I have the following but it looks awful:
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\begin{document}
\section{Introduction}
\bigskip
\smallskip
\subsection{Definitions and prerequisites}
Let us recall some basic knowledge, we begin by giving some definitions.
\medskip
\begin{definition}
A group is a set $G$ together with a binary operation $*$ on $G$ satisfying
the following properties:
\begin{enumerate}[label=(G\arabic*),series=group]
\item Closure: $\forall x,y \in G, x * y \in G$.
\item Associativity: $\forall x,y, z \in G, (x * y) * z = x * (y * z)$.
\item Identity: There is an element $e \in G$ such that $e * x = x * e = x$ for all $x \in G$.
\item Inverses: For any $x \in G$ there is an element $y \in G$ such that $x * y = y * x = e$.
\end{enumerate}
\end{definition}
\begin{definition}
A group $G$ is called an abelian group if the following axiom is satisfied:
\begin{enumerate}[label=(G\arabic*),resume=group]
\item Commutativity: $\forall x,y \in G, x * y = y * x$.
\end{enumerate}
\end{definition}
\medskip
Let $V$ be a vector space over the field $\mathbb{C}$ (unless stated otherwise) of complex numbers and let $GL(V)$ be the group of \textit{isomorphisms} of $V$ onto itself. An element $\alpha$ of $GL(V)$ is, by definition, a linear mapping of $V$ into $V$ which has an inverse $\alpha^{-1}$; this inverse is linear. When $V$ has a finite basis $(e_i)$ of $n$ elements, each linear map $\alpha: V \rightarrow V$ is defined by a square matrix $(\alpha_{ij})$ of order $n$. The coefficients $\alpha_{ij}$ are complex numbers; they are obtained by expressing the images $\alpha(e_j)$ in terms of the basis $(e_i)$:
\smallskip
$$\alpha(e_j)=\sum_i \alpha_{ij}e_i$$
\medskip
Saying that $\alpha$ is an isomorphism is identical to saying that the determinant $det(\alpha) = det(\alpha_{ij})$ of $\alpha$ is non zero. The group $GL(V)$ - the general linear group on $V$ is thus the group of \textit{invertible (or non-singular) square matrices of order $n$}.\\
\begin{definition}Suppose now $G$ is a finite group, with identity element $1$ and with composition $(s,t) \rightarrow st$. A linear representation of $G$ in $V$ is a homomorphism $\rho$ from the group $G$ into the group $GL(V)$(i.e. $\rho:G \rightarrow GL(V)$). In other words, we associate with each elements $s \in G$ and element $\rho(s)$ of $GL(V)$ in such a way that we have the equality:
$$\rho(st)=\rho(s)\cdot\rho(t)\;\; for\;\; s,t \in G. $$
Often we will use $\rho_s$ as an alternate to $\rho(s)$. We notice the above formula implies the following:
$$\rho(1)=1, \;\; \rho(s^{-1})=\rho(s)^{-1}.$$
\end{definition}
\medskip\
\\When $\rho$ is given, we simply say, by abuse of language, $V$ is a \textit{representation} of $G$ (or formally $V$ is a \textit{representation space} of $G$). Let us impose the condition that we are only to consider representations of finite degree, that is, in \textit{finite-dimensional} vector spaces; and these will almost always be vector spaces over $\mathbb{C}$. Therefore, to avoid repetition, let us agree to use the term "representation" to mean representation of finite degree over $\mathbb{C}$, unless stated otherwise. This isn't a strict limitation as for almost all applications we are interested in finite number of elements $x_i$ of $V$ hence we can find a a \textit{subrepresentation}.\\
Suppose now that $G$ has finite dimension, and let $n$ be its dimension; we say also that $n$ is the degree of the representation. Let $(e_i)$ be a basis of V, and let $R_s$ be the matrix of $\rho_s$ with respect to this basis. We have:\\
$$det(R_s)\neq 0, \;\;\;\;\;\;\;\; R_{st} = R_s \cdot R_t \;\;\;\; if\; s,t \in G$$\\
Let $r_{ij}(s)$ denote the coefficients of the matrix $R_s$, the second formula becomes:
$$r_{ik}(st)= \sum_j r_{ij}(s)\cdot r_{jk}(t)$$
Conversely, given the invertible matrices $R_s = (r_{ij}(s))$ satisfying the preceding identities, there is a corresponding linear representation of $\rho$ of $G$ in $V$; this is what we mean by giving a representation in matrix form.
\begin{definition}
let $\rho$ and $\rho\,'$ be two representations of the same group $G$ in vector spaces $V$ and $V\,'$. These representations are said to be similar (or isomorphic) if there exits a linear homomorphism $\tau: V \rightarrow V\,'$ which "transforms" $\rho$ to $\rho\,'$, that is which satisfies the identity:\
$$\tau \circ \rho(s) = \rho(s)\,'\circ \tau \;\;\;\;\; \forall s \in G$$
\end{definition}
When $\rho$ and $\rho\,'$ are given in matrix form by $R_s$ and $R_s\,'$, this means that that there exists an invertible matrix $T$ such that:
\smallskip
$$T\cdot R_s = R_s\,' \cdot T \;\;\;\;\; \forall s \in G$$
\subsection{Basic examples}
\end{document}
what must I do to remove the spacing in the last line?
$$,\rm,\caletc should not appear in any latex document written this century (\rmand\calhave not been defined by default since 1993) – David Carlisle Dec 11 '18 at 12:40proofand\begin{document}and remove all the\\and\...skipcommands and\;commands in the document, LaTeX can make beautiful documents but if you instruct it to make documents with arbitrary inconsistent spacing then you will find that it does what you ask. – David Carlisle Dec 11 '18 at 14:04;\; for\;\; s,t \in G. $$you have setforin math italic, which is designed to make adjacent letters not look like a word, but like a product of variables, and for the same reason the spaces in the input are ignored. try\text{ for }– David Carlisle Dec 11 '18 at 14:20