I am reading Coxeters "Generators and Relations for discrete goups". There is something in the very begin where I am struggling:
A group $G$ shall be defined by a set of $s$ relations $g_k$ between its $m$ generator elements $\{S_1,\cdots,S_m\}$ and the identity element $E$
\begin{align}
g_1(S_1,\dots,S_m) & = E \\
g_1(S_2,\dots,S_m) & = E \\
\cdots \tag{1}\\
g_s(S_2,\dots,S_m) & = E \\
\end{align}
and the group $G'$ shall be defined by the set of generators $\{R_1,\cdots,R_m\}$ which fulfill the same relations (1) and another $r$ relations:
\begin{align}
g_1(R_1,\dots,R_m) & = E \\
g_1(R_1,\dots,R_m) & = E \\
\cdots \tag{2}\\
g_s(R_1,\dots,R_m) & = E \\
\cdots \\
g_{s+r}(R_1,\dots,R_m) & = E \\
\end{align}
Now obviously the map
$$ S_i \mapsto R_i \;\;\; (i=1,…,m)$$ defines a group homomorphism from $G\to G'$ and all group elements
$$ g_k(R_1,\dots,R_m) \;\;\; (k=s+1,…,s+r)\tag{3}$$ correspond to the identity element $E$ in $G'$.
To me it seems that these elements form the kernel of the group homomorphism.
So I don't understand the next step in the argument in the book, it says
"the kernel of the homomorphism is the normal subgroup
$$ N \simeq \{W^{-1}g_k(S_1,\dots,S_m) W\} \;\;\; (k=s+1,…,s+r) \tag{4}$$
where $W$ runs through all the elements of $G$. In fact, $N$ is the smallest normal subgroup of $G$ that contains the elelemts (3), and it follows that
$$ G'\simeq G/N."$$
I somehow can't wrap my head around that conjugation loop over all elements how does it generate the normal subgroup.
I expect that it must be connected with the circumstance that iff $N$ is normal then
\begin{align}
WN & = NW\\
W^{-1}NW & = N = WNW^{-1}
\end{align}
must hold for all $W\in G$. But I still can't see clearly how it connects to construction (4). Possibly its very simple but I am just hanging here. Would be grateful for any hints.
Best Answer
Importantly, the kernel of a group homomorphism is always a normal subgroup, so that it's closed under conjugations: if $f(x)=e$, then $f(gxg^{-1})=f(g)\cdot e\cdot f(g)^{-1}=e$.
For group presentations, it corresponds to a natural property of equality: if the word $x$ is evaluated as the identity element, meaning $x=e$ in the presented group, then for every other word $g$, we must also have $gxg^{-1}=geg^{-1}=e$.
In our example with $\varphi:G\to G'$, we obviously have $g_k(S_1,\dots, S_m)\in \ker \varphi$ if $k>s$, so the normal subgroup $N$ generated by these elements - which is the same as the subgroup generated by all conjugations of these -, is fully included in $\ker\varphi$.
Moreover, it turns out that $N=\ker\varphi$, and so $G'\cong G/N$.