# Topology/Free group and presentation of a group
## Free monoid spanned by a set

Let $V$ be a vector space and $v_{1},\ldots ,v_{n}$ be a basis of $V$. Given any vector space $W$ and any elements $w_{1},\ldots ,w_{n}\in W$, there is a linear transformation $\varphi :V\rightarrow W$ such that $\forall i\in \{1,\ldots ,n\},\,\varphi (v_{i})=w_{i}$. One could say that this happens because the elements $v_{1},\ldots ,v_{n}$ of a basis are not "related" to each other (formally, they are linearly independent). Indeed, if, for example, we had the relation $v_{1}=\lambda v_{2}$ for some scalar $\lambda$ (and then $v_{1},\ldots ,v_{n}$ wasn't linearly independent), then the linear transformation $\varphi$ could not exist.
Let us consider a similar problem with groups: given a group $G$ spanned by a set $X=\{x_{i}:i\in I\}\subseteq G$ and given any group $H$ and any set $Y=\{y_{i}:i\in I\}\subseteq H$, does there always exist a group morphism $\varphi :G\rightarrow H$ such that $\forall i\in I,\,\varphi (x_{i})=y_{i}$? The answer is no. For example, consider the group $G=\mathbb {Z} _{n}=\mathbb {Z} /n\mathbb {Z}$ which is spanned by the set $X=\{1\}$, the group $H=\mathbb {R}$ (with the adition operation) and the set $Y=\{2\}$. If there exists a group morphism $\varphi :\mathbb {Z} _{n}\rightarrow \mathbb {R}$ such that $\varphi (1)=2$, then $n2=n\varphi (1)=\varphi (n\,1)=\varphi (0)=0$, which is impossible. But if instead we had choose $G=\mathbb {Z}$, then such a group morphism does exist and it would be given by $\varphi (t)=2t$. Indeed, given any group $H$ and any $y\in H$, we have the group morphism $\varphi :\mathbb {Z} \rightarrow H$ defined by $\varphi (t)=y^{t}$ (in multiplicative notation) that verifies $\varphi (1)=y$. In a way, we can think that this happens because the elements of the set $X=\{1\}\subseteq \mathbb {Z}$ (that spans $\mathbb {Z}$) don't verify relations like $nx=1$ (like $\mathbb {Z} _{n}$) or $xy=yx$. So, it seems that $\mathbb {Z}$ is a group more "free" that $\mathbb {Z} _{n}$.
Our goal in this section will be, given a set $X$, build a group spanned by the set $X$ such that it will be the most "free" possible, in the sense that it doesn't have to obey relations like $x^{n}=1$ or $xy=yx$. To do so, we begin by constructing a "free" monoid (in the same sense). Informally, this monoid will be the monoid of the words written with the letters of the alphabet $X$, where the identity will be the word with no letters (the "empty word"), and the binary operation of the monoid will be concatenation of words. The notation $x_{1}\ldots x_{n}$ that we will use for the element of this monoid meets this idea that the elements of this monoid are the words $x_{1}\ldots x_{n}$ where $x_{1},\ldots ,x_{n}$ are letters of the alphabet $X$. Here is the definition of this monoid.
Definition Let $X$ be a set.
* We denote the $n$-tuples $(x_{1},\ldots ,x_{n})$ with $x_{i}\in X$ and $n\in \mathbb {N}$ by $x_{1}\ldots x_{n}$.
* We denote $$, that is $(x_{1},\ldots ,x_{n})$ with $n=0$, by $1$.
* We denote by $FM(X)$ the set $\{x_{1}\ldots x_{n}:n\in \mathbb {N} ,x_{i}\in X\}$.
* We define in $FM(X)$ the concatenation operation $*$ by $x_{1}\ldots x_{m}*y_{1}\ldots y_{n}=x_{1}\ldots x_{m}y_{1}\ldots y_{n}$.
Next we prove that this monoid is indeed a monoid. It's an easy to prove result, we need to show associativity of $*$ and that $1*x=x*1=x$.
Proposition $(FM(X),*)$ is a monoid with identity $1$.
Proof The operation $*$ is associative because, given any $x_{1}\ldots x_{m},y_{1}\ldots y_{n},z_{1}\ldots z_{p}\in FM(X)$, we have
    $(x_{1}\ldots x_{m}*y_{1}\ldots y_{n})*z_{1}\ldots z_{m}$
    $=x_{1}\ldots x_{m}y_{1}\ldots y_{n}*z_{1}\ldots z_{m}$
    $=x_{1}\ldots x_{m}y_{1}\ldots y_{n}z_{1}\ldots z_{m}$
    $=x_{1}\ldots x_{m}*(y_{1}\ldots y_{n}z_{1}\ldots z_{m})$
    $=x_{1}\ldots x_{m}(y_{1}\ldots y_{n}*z_{1}\ldots z_{m})$.
It's obvious that $(FM(X),*)$ has the identity $1$ as $1*x_{1}\ldots x_{n}=x_{1}\ldots x_{n}$ by the definition of $1$ and $*$. $\square$
Following the idea that the monoid $(FM(X),*)$ is the most "free" monoid spanned by $X$, we will call it the free monoid spanned by $X$.
Definition Let $X$ be a set. We denote the free monoid spanned by $X$ by $(FM(X),*)$.
Examples
* Let $X=\{x\}$. Then $FM(X)=\{1,x,xx,xxx,\ldots \}$ and, for example, $xx*xxx=xxxxx$.
* Let $X=\{x,y,z\}$. Then $1,x,y,z,xxx,yxz,xyzzz\in FM(X)$ and, for example, $xxx*yxz=xxxyxz$.

## Free group spanned by a set

Now let us construct the more "free" group spanned by a set $X$. Informally, what we will do is insert in the monoid $FM(X)$ the inverse elements that are missing in it for it to be a group. In a more precise way, we will have a set ${\bar {X}}$ equipotent to $X$, choose a bijection from $X$ to ${\bar {X}}$ and in this way achieve a "association" between the elements of $X$ and the elements of ${\bar {X}}$. Then we face $x_{1}\ldots x_{n}\in FM(X)$ (with $x_{1},\ldots ,x_{n}\in X$) as having the inverse element ${\overline {x_{n}}}\ldots {\overline {x_{1}}}$ (with ${\overline {x_{1}}},\ldots ,{\overline {x_{n}}}\in X$), where $x_{1},\ldots ,x_{n}\in X$ is is associated with ${\overline {x_{1}}}\ldots {\overline {x_{n}}}$, respectively. Let us note that the order of the elements in ${\overline {x_{n}}}\ldots {\overline {x_{1}}}$ is "reversed" because the inverse of the product $x_{1}\ldots x_{n}=x_{1}*\cdots *x_{n}$ must be $x_{n}^{-1}*\cdots *x_{1}^{-1}$, and the $x_{1}^{-1},\ldots ,x_{n}^{-1}$ are, respectively, ${\overline {x_{1}}},\ldots ,{\overline {x_{n}}}$. The way we do that ${\overline {x_{n}}}\ldots {\overline {x_{1}}}$ be the inverse of $x_{1}\ldots x_{n}$ is to take a congruence relation $R$ that identifies $x_{1}\ldots x_{n}{\overline {x_{n}}}\ldots {\overline {x_{1}}}$ with $1$, and pass to the quotient $FM(X\cup {\bar {X}})$ by this relation (defining then, in a natural way, the binary operation of the group, $[u]_{R}\star [v]_{R}=[u*v]_{R}$). By taking the quotient, we are formalizing the intuitive idea of identifying $x_{1}\ldots x_{n}{\overline {x_{n}}}\ldots {\overline {x_{1}}}$ with $1$, because in the quotient we have the equality $[x_{1}\ldots x_{n}{\overline {x_{n}}}\ldots {\overline {x_{1}}}]_{R}=[1]_{R}$. Let us give the formal definition.
Definition Let $X$ be a set. Let us take another set ${\overline {X}}$ equipotent to $X$ and disjoint from $X$ and let $f:X\rightarrow {\overline {X}}$ be a bijective application.
* For each $x\in X$ let us denote $f(x)$ by ${\bar {x}}$, for each $x\in {\overline {X}}$ let us denote $f^{-1}(x)$ by ${\bar {x}}$ and for each $x_{1}\ldots x_{n}\in FM(X\cup {\overline {X}})$ let us denote ${\overline {x_{n}}}\ldots {\overline {x_{1}}}$ by ${\overline {x_{1}\ldots x_{n}}}$.
* Let $R$ be the congruence relation of $FM(X\cup {\overline {X}})$ spanned by $G=\{(u*{\bar {u}},1):u\in X\cup {\overline {X}}\}$, this is, $R$ is the intersection of all the congruence relations in $FM(X\cup {\overline {X}})$ wich have $G$ as a subset. We denote the quotient set $FM(X\cup {\overline {X}})/R$ by $FG(X)$.
Frequently, abusing the notation, we represent an element $[u]_{R}\in FG(X)$ simply by $u$.
Because the operation $[u]_{R}\star [v]_{R}=[u*v]_{R}$ that we want to define in $FM(X\cup {\bar {X}})/R$ is defined using particular represententes $u$ and $v$ of the equivalence classes $[u]_{R}$ and $[v]_{R}$, a first precaution is to verify that the definition does not depend on the chosen representatives. It's an easy verification.
Lemma Let $X$ be a set. It is well defined in $FG(X)$ the binary operation $\star$ by $[u]_{R}\star [v]_{R}=[u_{r}*v_{r}]_{R}$ (where $R$ is the congruence relation of the previous definition).
Proof Let $u,u',v,v'\in FM(X)$ be any elements such that $[u]_{R}=[u']_{R}$ and $[v]_{R}=[v']_{R}$, this is, $uRu'$ and $vRv'$. Because $R$ is a congruence relation in $FM(X\cup {\bar {X}})$, we have $u*vRu'*v'$, this is, $[u*v]_{R}=[u'*v']_{R}$. $\square$
Because the definition is valid, we present it.
Definition Let $X$ be a set. We define in $FG(X)$ the binary operation $\star$ by $[u]_{R}\star [v]_{R}=[u_{r}*v_{r}]_{R}$.
Finally, we verify that the group that we constructed is indeed a group.
Proposition Let $X$ be a set. $(FG(X),\star )$ is a group with identity $[1]_{R}$ and where $\forall [u]_{R}\in FG(X),\,{[u]_{R}}^{-1}=[{\bar {u}}]_{R}$.
Proof
* $(FG(X),\star )$ is associative because $\forall [u]_{R},[v]_{R},[w]_{R}\in FG(X),\,([u]_{R}\star [v]_{R})\star [w]_{R}=([u*v]_{R})\star [w]_{R}=[([u]_{R}*[v]_{R})*w]_{R}=$ $[u*(v*w)]_{R}=[u]_{R}\star [v*w]_{R}=[u]_{R}\star ([v]_{R}\star [w]_{R}).$
* Let us see that $[1]_{R}$ is the identity $(FG(X),\star )$. Let $[u]_{R}\in FG(X)$ be any element. We have $[u]_{R}\star [1]_{R}=[u*1]_{R}=[u]_{R}$ and, in the same way, $[1]_{R}\star [u]_{R}=[u]_{R}$.
* Let $[u]_{R}\in FG(X)$ be any element and let us see that $[u]_{R}\star [{\bar {u}}]_{R}=[1]_{R}$. We have $[u]_{R}\star [{\bar {u}}]_{R}=[u*{\bar {u}}]_{R}$ and, by definition of $R$, $u*{\bar {u}}R1$, this is, $[u*{\bar {u}}]_{R}=[1]_{R}$, therefore $[u]_{R}\star [{\bar {u}}]_{R}=[1]_{R}$ and, in the same way, $[{\bar {u}}]_{R}\star [u]_{R}=[1]_{R}$. $\square$
In the same way that we did with the free monoid, we will call free group spanned by the set $X$ to the more "free" group spanned by this set.
Definition Let $X$ be a set. We call free group spanned by $X$ to $(FG(X),\star )$.
Example Let $X=\{x\}$. Let us choose any set ${\bar {X}}=\{y\}$ disjoint (and equipotent) of $X$. Let $f:X\rightarrow {\bar {X}}$ be any (in fact, the only) bijective application of $X$ in ${\bar {X}}$. Then we denote $f(x)=y$ by ${\bar {x}}$ and we denote $f^{-1}(y)=x$ by ${\bar {y}}$. We regard $x$ and $y$ as inverse elements. Let $R$ be the congruence relation of $FM(X\cup {\bar {X}})$ spanned by $G=\{(1,1),(x{\bar {x}},1),(xx{\bar {x}}{\bar {x}},1),\ldots \}$. $FG(X)=FM(\{x,{\bar {x}}\})/R$ is the set of all "words" written in the alphabet $\{[x]_{R},[{\bar {x}}]_{R}\}$. For example, $[1]_{R},[x]_{R},[{\bar {x}}]_{R},[xx{\bar {x}}xx]_{R}\in FG(X)$.
We have $G\subseteq R$ and, for example, $(xx{\bar {x}},1)\in R$, because $(x{\bar {x}},1)\in G\subseteq R$ (therefore $x{\bar {x}}R1$) and because $R$ is a congruence relation, we can "multiply" both "members" of the relation $x{\bar {x}}R1$ by $x$ and obtain $xx{\bar {x}}Rx$. We see $x{\bar {x}}R1$ as meaning that in $FG(X)$ We have $xx{\bar {x}}=x$ (more precisely, $[xx{\bar {x}}]_{R}=[x]_{R}$), and we think in this equality as being the result of one $x$ "cut out" with ${\bar {x}}$ in $xx{\bar {x}}$.
Given $u\in FM(X\cup {\bar {X}})$, let us denote the exact number of times that the "letter" $x$ appears in $u$ by $|u|_{x}$ and let us denote the exact number of times the "letter" ${\bar {x}}$ appears in $u$ by $|u|_{\bar {x}}$. Then "cutting" $x$'s with ${\bar {x}}$'s it remains a reduced word word with $|u|_{x}-|u|_{\bar {x}}$ times the letter $x$ (if $|u|_{x}-|u|_{\bar {x}}<0$, let us us consider that there aren't any letters $x$ and remains $-(|u|_{x}-|u|_{\bar {x}})$ times the letter ${\bar {x}}$). Let us denote $|u|_{x}-|u|_{\bar {x}}$ by $|u|_{x-{\bar {x}}}$. We have
* $[u]_{R}=[v]_{R}$ if and only if $|u|_{x-{\bar {x}}}=|v|_{x-{\bar {x}}}$ and
* $\forall [u]_{R},[v]_{R}\in FG(X),\,|uv|_{x-{\bar {x}}}=|u|_{x-{\bar {x}}}+|v|_{x-{\bar {x}}}$.
In this way, each element $[u]_{R}\in FG(X)$ is determined by the integer number $|u|_{x-{\bar {x}}}$ and the product $\star$ of two elements $[u]_{R},[v]_{R}\in FG(X)$ correspondent to the sum of they associated integers numbers $|u|_{x-{\bar {x}}}$ and $|v|_{x-{\bar {x}}}$. Therefore, it seems that the group $(FG(X),\star )$ is "similar" to $(\mathbb {Z} ,+)$. indeed $(FG(X),\star )$ is isomorph to $(\mathbb {Z} ,+)$ and the application $|\cdot |_{x-{\bar {x}}}:FG(X)\rightarrow \mathbb {Z}$ is a group isomorphism.

## Presentation of a group

Informally, it seems that $\mathbb {Z} _{n}$ is obtained from the "free" group $\mathbb {Z}$ imposing the relation $nx=1$. Let us try formalize this idea. We start with a set $X$ that spans a group $G$ that que want to create and a set of relations $R$ (such as $x^{n}=1$ or $xy=yz$) that the elements of $G$ must verify and we obtain a group $G/R$ spanned by $G$ and that verify the relations of $R$. More precisely, we write each relation $u=v$ in the form $uv^{-1}=1$ (for example, $xy=yx$ is written in the form $xyx^{-1}y^{-1}=1$) and we see each $uv^{-1}$ as a "word" of $FG(X)$. Because $R$ doesn't have to be a normal subgroup of $G$, we can not consider the quotient $FG(X)/R$, so we consider the quotient $FG(X)/N$ where $N$ is the normal subgroup of $FG(X)$ spanned by $R$. In $G/N$, we will have $uv^{-1}N=1N$, which we see as meaning that in $G/N$ the elements $uv^{-1}$ and $1$ are the same. In this way, $FG(X)/N$ will verify all the relations that we want and will be spanned by $X$ (more precisely, by $\{xN:x\in X\}$). Let us formalize this idea.
Definition Let $G$ be a group. We call presentation of $G$, and denote by $<X:R>$, to a ordered pair $(X,R)$ where $X$ is a set, $R\subseteq FG(X)$ and $G\simeq FG(X)/N$, where $N$ is the normal subgroup of $FG(X)$ spanned by $R$. Given a presentation $<X:R>$, we call spanning set to $X$ and set of relations to $R$.
Let us see some examples of presentations of the free group $FG(X)$ and the groups $\mathbb {Z} _{n}$, $\mathbb {Z} \oplus \mathbb {Z}$, $\mathbb {Z} _{m}\oplus \mathbb {Z} _{n}$ and $S_{3}$. We also use the examples to present some common notation and to show that a presentation of a group does not need to be unique.
Examples
* Let $X$ be a set. $<X:{\emptyset }>$ is a presentation of $FG(X)$ because $FG(X)\simeq FG(X)/\{1\}$, where $\{1\}$ is the normal subgroup of $FG(X)$ spanned by $\emptyset$. In particular, $<\{x\}:\emptyset >$ is a presentation of $(\mathbb {Z} ,+)\simeq FG(\{x\})$, more commonly denoted by $<x:\emptyset >$. Another presentation of $(\mathbb {Z} ,+)$ is $<\{x,y\}:\{xy^{-1}\}>$, more commonly denoted by $<x,y:xy^{-1}>$. Informally, in the presentation $<x,y:xy^{-1}>$ we insert a new element $y$ in the spanning set, but we impose the relation $xy^{-1}=1$, this is, $x=y$, which is the same as having not introduced the element $y$ and have stayed by the presentation $<x:\emptyset >$.
* Let $X=\{x\}$. $<X:\{x^{n}\}>$ (where $x^{n}=x\star \cdots \star x\in FG(X)$ $n$ times) is a presentation of $\mathbb {Z} _{n}$. Indeed, the subgroup of $FG(X)$ spanned by $\{x^{n}\}$ is $N=\{\ldots ,{\bar {x}}^{2n},{\bar {x}}^{n},1,x^{n},x^{2n},\ldots \}\simeq n\mathbb {Z}$ and $FG(X)\simeq \mathbb {Z}$, therefore$\mathbb {Z} _{n}=\mathbb {Z} /n\mathbb {Z} \simeq FG(X)/N$. Is more common to denote $<\{x\}:\{x^{n}\}>$ by $<x:x^{n}>$.
* Let $X=\{x,y\}$ (with $x$ and $y$ distinct) and $R=\{xyx^{-1}y^{-1}\}$. $<X:R>$ is a presentation of $\mathbb {Z} \oplus \mathbb {Z}$. Informally, what we do is impose comutatibility in $FG(X)$, this is, $xy=yx$, this is, $xyx^{-1}y^{-1}=1$, obtaining a group isomorph to $\mathbb {Z} \oplus \mathbb {Z}$. It's more usually denote $<\{x,y\}:\{xyx^{-1}y^{-1}\}>$ by $<x,y:xyx^{-1}y^{-1}>$.
* Let $X=\{x,y\}$ and $R=\{xyx^{-1}y^{-1},x^{m},y^{n}\}$. $<X,R>$ is a presentation of $\mathbb {Z} _{m}\times \mathbb {Z} _{n}$. Informally, what we do is impose commutability in the same way as in the previous example, and we impose $x^{m}=1$ and $x^{n}=1$ to obtain $\mathbb {Z} _{m}\times \mathbb {Z} _{n}$ instead $\mathbb {Z} \oplus \mathbb {Z}$. It's more common denote$<\{x,y\}:\{xyx^{-1}y^{-1},x^{m},y^{n}\}>$ by $<x,y:xyx^{-1}y^{-1},x^{m},y^{n}>$.
* $<\{a,b,c\}:\{aa,bb,cc,abac,cbab\}>$, more commonly written $<a,b,c:a^{2},b^{2},c^{2},abac,cbab>$, is a presentation of $S_{3}$, the group of the permutations of $\{1,2,3\}$ with the composition of applications. To verify this, one can verify that any group with presentation $<a,b,c:a^{2},b^{2},c^{2},abac,cbab>$ as exactly six elements $id$, $a$, $b$, $c$, $a$, $ab$ and $ac$, and that the multiplication of this elements results in the following Cayley table that is equal to the Cayley table of $S_{3}$. Just to give an idea how this can be achieved, a group with presentation $<a,b,c:a^{2},b^{2},c^{2},abac,cbab>$ as exactly the elements $id$, $a$, $b$, $c$, $a$, $ab$ and $ac$ because none of this elements are the same (the relations $a^{2}=b^{2}=c^{2}=abac=cbab=1$ don't allow us to conclude that two of the elements are equal) and because "another" elements like $bc$ are actually one of the previous elements (for example, from $cbab=id$ we have $cb=ab$, and taking inverses of both members, we have $b^{-1}c^{-1}=b^{-1}a^{-1}$, which, using $a^{2}=b^{2}=c^{2}=id$, this is, $a=a^{-1}$, $b=b^{-1}$ and $c=c^{-1}$, results in $bc=ba$). Then, using the relations of the presentation, one can compute the Cayley table. For example, $a(ab)=b$ because we have the relation $a^{2}=1$. Another example: we have $b(ac)=a$ because we can multiply both members of the relation $abac=id$ by $a$ and then use $a^{2}=id$. One could have suspected of this presentation by taking $a=(1\ 2)$, $b=(1\ 3)$ and $c=(2\ 3)$ and then, trying to construct the Cayley table of $S_{3}$, found out that it was possible if one know that $a^{2}=b^{2}=c^{2}=abac=cbab=1$.
| $\times$ | $id$ | $a$ | $b$ | $c$ | $ab$ | $ac$ |
| $id$ | $id$ | $a$ | $b$ | $c$ | $ab$ | $ac$ |
| $a$ | $a$ | $id$ | $ab$ | $ac$ | $b$ | $c$ |
| $b$ | $b$ | $ac$ | $id$ | $ab$ | $c$ | $a$ |
| $c$ | $c$ | $ab$ | $ac$ | $id$ | $a$ | $b$ |
| $ab$ | $ab$ | $c$ | $a$ | $b$ | $ac$ | $id$ |
| $ac$ | $ac$ | $b$ | $c$ | $a$ | $id$ | $ab$ |
It's natural to ask if all groups have a presentation. The following theorem tell us that the answer is yes, and it give us a presentation.
Theorem Let $(G,\times )$ be a group.
* The application $\varphi :FG(G)\rightarrow G$ defined by $\varphi ([x_{1}]_{R}\star \cdots \star [x_{n}]_{R})=x_{1}\times \cdots \times x_{n}$ (where $x_{1},\ldots ,x_{n}\in G$) is an epimorphism of groups.
* $<G:{\textrm {ker}}\,\varphi >$ is a presentation of $(G,\times )$.
Proof
* $\varphi$ is well defined because every element of $FG(X)$ as a unique representations in the form $[x_{n}]\star \cdots [x_{n}]_{R}$ with $x_{1},\ldots ,x_{n}\in G$, with the exception of $[1]_{R}$ appears several times in the representation, but that doesn't affect the value of $x_{1}\times \cdots \times x_{n}$ Let $[x_{1}]_{R}\star \cdots \star [x_{m}]_{R},[y_{1}]_{R}\star \cdots \star [y_{n}]_{R}\in FG(X)$ be any elements, where $x_{1},\ldots ,x_{m},y_{1},\ldots ,y_{n}\in G$. We have $\varphi (([x_{1}]_{R}\star \cdots \star [x_{m}]_{R})\star ([y_{1}]_{R}\star \cdots \star [y_{n}]_{R}))=(x_{1}\times \cdots \times x_{m})\times (y_{1}\times \cdots \times y_{n})=$ $\varphi ([x_{1}]_{R}\star \cdots \star [x_{m}]_{R})\times \varphi ([y_{1}]_{R}\star \cdots \star [y_{n}]_{R})$, therefore $\varphi$ is a morphism of groups. Because $\forall x\in G,\,\varphi ([x]_{R})=x$, then $G$ is a epimorphism of groups.
* Using the first isomorphism theorem (for groups), we have $FG(G)/{\textrm {ker}}\,\varphi \simeq \varphi (G)=G$, therefore $<G:{\textrm {ker}}\,\varphi >$ is a presentation of $(G,\times )$. $\square$
The previous theorem, despite giving us a presentations of the group $G$, doesn't give us a "good" presentation, because the spanning set $G$ is usually much larger that other spanning sets, and the set of relations $\mathrm {ker} \,\varphi$ is also usually much larger then other sufficient sets of relations (it is even a normal subgroup of $FG(G)$, when it would be enough that it span an appropriated normal subgroup).