Kernel is the normal closure

Question

Let $f \colon G \mapsto \mathbb{Z}$ be an epimorphism and suppose that $\{g_1,\dots,g_n\}$ is a finite generating set for $G$. Let $x\in G$ such that $f(x)=1$. Then, for each $i\in \{1,\dots, n\}$ there is $m_i \in \mathbb{Z}$ such that $g_i x^{m_i}\in \ker f$.

In those conditions, is it true that $\ker f$ is normally generated in $G$ by $g_1 x^{m_1},\dots,g_n x^{m_n}$? That is, is it true that $\ker f= \langle\langle g_1 x^{m_1},\dots,g_n x^{m_n}\rangle\rangle_G$?

Moreover, since $G= \ker f \langle x \rangle$, is $\{ (g_i x^{m_i})^{x^k} \mid k\in \mathbb{Z}, i\in \{1,\dots,n\} \}$ a generating set for $\ker f$?

It is a property that I have used a lot and now I am doubting it... Thank you!

Answer 1

Let $H=\langle g_1x^{m_1},\ldots,g_nx^{m_n}\rangle$, and let $M$ be the subgroup generated by conjugates of elements of $H$ by powers of $x$. To show that $M$ is normal in $G$, because the $g_i$ generate $G$, it is enough to show that if $u\in M$, then $u^{g_i}\in M$ and $u^{g_i^{-1}}\in M$ for each $i$. Note that $M$ is closed under conjugation by powers of $x$.

Indeed, we have $$ u^{g_i} = u^{g_ix^{m_i}x^{-m_i}}= (u^{g_ix^{m_i}})^{x^{-m_i}}.$$ Because $g_ix^{m_i}\in H\subseteq M$, and $u\in M$, then $u^{g_ix^{m_i}}\in M$. And since $M$ is closed under conjugation by powes of $x$, it follows that $u^{g_i}\in M$.

For inverses, we have $$u^{g_i^{-1}} = u^{x^{m_i}x^{-m_i}g_i^{-1}} = (u^{x^{m_i}})^{x^{-m_i}g_i^{-1}}.$$ Since $u\in M$ and $M$ is closed under conjugation by powers of $x$, we have $u^{x^{m_i}}\in M$. And since $x^{-m_i}g_i^{-1}\in H\leq M$, then $u^{g_i^{-1}}\in M$, as desired.

This shows that $M\triangleleft G$, and therefore that the normal closure of $H$ in $G$ is contained in $M$. Since $M$ is trivially contained in the normal closure, we get that $M$ is indeed the normal closure of $H$.

Clearly $M$ is contained in the kernel, since each of the generators of $H$ is contained in the kernel. Suppose that $z\in\ker f$, and write $$z = g_{i_1}^{\epsilon_1}\cdots g_{i_k}^{\epsilon_k},$$ with $\epsilon_j\in\{1,-1\}$. Then because $f(g_{i_k}) = -m_{i_k}$, we must have that $$\epsilon_1m_{i_1}+\epsilon_2m_{i_1}+\cdots+\epsilon_km_{i_k}=0.$$ Note also that $x^{m_i}g_i = x^{m_i}(g_ix^{m_i})x^{-m_i}\in M$, so both $g_ix^{m_i}$ and $g_i^{-1}x^{-m_i}\in M$.

Thus we have: $$\begin{align*} z &= g_{i_1}^{\epsilon_1}\cdots g_{i_k}^{\epsilon_k}\\ &= (g_{i_1}^{\epsilon_1} x^{\epsilon_1m_{i_1}})\cdot x^{-\epsilon_1m_{i_1}} (g_{i_2}^{\epsilon_2}x^{\epsilon_2m_{i_2}})x^{\epsilon_1m_{i_1}}\\ &\qquad\cdot x^{-\epsilon_1m_{i_1}-\epsilon_2m_{i_2}}(g_{i_3}^{\epsilon_3}x^{\epsilon_3m_{i_3}})x^{\epsilon_1m_{i_1}+\epsilon_2m_{i_2}}\cdot x^{-\epsilon_1m_{i_1}-\epsilon_2m_{i_2}-\epsilon_3m_{i_3}}\\ &\qquad \cdots x^{-\epsilon_1m_{i_1}-\cdots-\epsilon_{k-1}m_{i_{k-1}}} (g_{i_k}^{\epsilon_k}x^{\epsilon_km_{i_k}})x^{\epsilon_1m_{i_1}+\cdots+\epsilon_{k-1}m_{i_{k-1}}}\\ &\qquad \cdot x^{-\epsilon_1m_{i_1}-\cdots - \epsilon_km_{i_k}}\\ &= (g_{i_1}^{\epsilon_1} x^{\epsilon_1m_{i_1}})\\ &\qquad \cdot x^{-\epsilon_1m_{i_1}} (g_{i_2}^{\epsilon_2}x^{\epsilon_2m_{i_2}})x^{\epsilon_1m_{i_1}}\\ &\qquad\cdot x^{-\epsilon_1m_{i_1}-\epsilon_2m_{i_2}}(g_{i_3}^{\epsilon_3}x^{\epsilon_3m_{i_3}})x^{\epsilon_1m_{i_1}+\epsilon_2m_{i_2}}\\ &\qquad \cdots x^{-\epsilon_1m_{i_1}-\cdots-\epsilon_{k-1}m_{i_{k-1}}} (g_{i_k}^{\epsilon_k}x^{\epsilon_km_{i_k}})x^{\epsilon_1m_{i_1}+\cdots+\epsilon_{k-1}m_{i_{k-1}}} \end{align*}$$ with the last equality because $\epsilon_1m_{i_1}+\cdots + \epsilon_km_{i_k}=0$. This is a product of conjugates of the generators of $H$, hence lies in $M$. Thus, $\ker(f)\subseteq M$, proving the equality.

Answer 2

The answer to both is "yes". Since the second clearly implies the first, I will concentrate on proving the second.

Thus, let $H$ be the subgroup of $G$ generated by $\{ (g_i x^{m_i})^{x^k} \}$. I claim that for every $y \in G$, we have $y x^{-f(y)} \in H$. To see this, consider the subset of $y \in G$ satisfying this condition. Then:

• For each $i$, $g_i x^{-f(g_i)} = g_i x^{m_i} \in H$, so each $g_i$ is in this set.
• Suppose $y$ and $y'$ are both in the set. Then $(yy') x^{-f(yy')} = y y' x^{-f(y')} x^{-f(y)} = y x^{-f(y)} (y' x^{-f(y')})^{x^{-f(y)}}$. Since $y x^{-f(y)}\in H$, $y' x^{-f(y')}\in H$, and it is easy to see that $H$ is closed under conjugation by $x^k$ for every $k$, this implies that $yy'$ is also in the set.
• Suppose $y$ is in the set. Then $y^{-1} x^{-f(y^{-1})} = y^{-1} x^{f(y)} = [(y x^{-f(y)})^{-1}]^{x^{f(y)}}$. Since $y x^{f(y)}\in H$, and again $H$ is closed under conjugation by $x^k$ for every $k$, this implies that $y^{-1}$ is also in the set.

In conclusion, the set is a subgroup of $G$ which contains each $g_i$, so it is all of $G$, completing the proof of the claim.

As a corollary, if $y \in \ker(f)$, then $y \in H$. Since it is clear that $H \subseteq \ker(f)$ also, we get that $H = \ker(f)$.