Created the Abstract Algebra theorems and definitions cheat sheet
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\section{Generators and Relations}
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\begin{definition}[Generators and Relations]
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Let $G$ be a group generated by some subset $A=\{a_1,a_2,\dots,a_n\}$ and let $F$ be the free group on $A$. Let $W = \{w_1,w_2,\dots,w_t\}$ be a subset of $F$ and let $N$ be the smallest normal subgruop of $F$ containing $W$. We say that $G$ is \textit{given by the generators $a_1,a_2,\dots,a_n$ and the relations $w_1=w_2=\dots=w_t=e$} if there is an isomorphism from $F/N$ onto $G$ that carries $a_iN$ to $a_i$.
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\noindent The notation for this situation is
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\[ G = \lr{a_1,a_2,\dots,a_n\ \vert\ w_1=w_2=\dots=w_t=e} \]
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\end{definition}
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\begin{theorem}[Dyck's Theorem (1882)]
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Let
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\[ G = \lr{a_1,a_2,\dots,a_n\ \vert\ w_1=w_2=\dots=w_t=e} \]
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and let
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\[ \overline{G}=\lr{a_1,a_2,\dots,a_n\ \vert\ w_1=w_2=\dots=w_t=w_{t+1}=\dots=w_{t+k}=e} \]
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Then $\overline{G}$ is a homomorphic image of $G$.
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\end{theorem}
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\begin{corollary}[Largest Group Satisfying Defining Relations]
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If $K$ is a group satisfying the defining relations of a finite group $G$ and $\abs{K} \geq \abs{G}$, then $K$ is isomorphic to $G$.
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\end{corollary}
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