Suppose we have two groups $G = \langle g_1, \cdots, g_r, \cdots g_n\rangle$, $H = \langle g_1, \cdots, g_r\rangle$ and a homomorphism $f: G \to H$ such that $f(g_i) = g_i$ if $i\leq r$ and $f(g_i) = 1$ otherwise. How can I show that $\ker f$ is the normal closure of $\{g_{r+1}, \cdots, g_n\}$?
I've shown that the normal closure is in the kernel, but I'm having trouble proving the other direction of inclusion. I feel like it can be done by the isomorphism theorems, but I don't know how exactly.
Best Answer
Note that because $H$ is a retract of $G$, then $\ker(f)\cap H=\{e\}$, and every element of $G$ can be written uniquely as $nh$ with $n\in\ker(f)$ and $h\in H$.
Now, let $K$ be the normal closure of $\langle g_{r+1},\ldots,g_n\rangle$. You've proven that $K\leq\ker(f)$. Note that $KH=G$, since $K$ is normal so $KH=\langle K,H\rangle$, and $K\cup H$ contains a generating set of $G$; and $K\cap H\leq \ker(f)\cap H = \{e\}$, so every element of $G$ can also be written uniquely as $kh$ with $k\in K$ and $h\in H$: if $kh=k'h'$, then $(k')^{-1}k = h'h^{-1}\in K\cap H = \{e\}$, so $k=k'$ and $h=h'$.
Let $n\in \ker(f)$. Then we can write $n=kh$ with $k\in K$ and $h\in H$; but because $ne=kh$, with $n,k\in \ker(f)$ and $e,h\in H$, then the uniqueness of the decomposition mean that $n=k$ and $h=e$. Thus, $n\in K$, proving that $\ker(f)\leq K$, yielding the reverse inclusion.