For which subgroups the transfer map kills a given element of a group?

Question

$\newcommand{\ab}{{\rm ab}} \newcommand{\ord}{{\rm ord}} $Let $G$ be a finite or profinite group. Consider the abelianized group $$G^\ab=G/G'$$ where $G'$ is the commutator subgroup of $G$.

Let $H\subset G$ be a subgroup of finite index. Consider the transfer (Verlagerung) map $$V_H\colon G^\ab\to H^\ab\,;$$ see https://en.wikipedia.org/wiki/Transfer_(group_theory).

Fix $g\in G$ and write $a=gG'\in G^\ab$. I think that if $V_H(a)=0\in H^\ab$, then $na=0\in G^\ab$ where $n=[G:H]$, because ${\rm Cor}\circ{\rm Res}=n$. Therefore, $$ \ord(a) \mid n$$ where $\ord(a)$ denotes the order of $a$ in the abelian group $G^\ab$. It follows that $$ \ord(a)\mid \gcd\nolimits_H[G:H]$$ where $\gcd_H$ denotes the greatest common divisor over the subgroups of finite index $H$ such that $V_H(a)=0$.

Question. For given $a\in G^\ab$, what are the relations between the order $\ord(a)$ and the possible indices of the subgroups $H$ such that $V_H(a)=0$? In particular,

(1) Can $\ord(a)$ be strictly smaller than $\gcd_H [G:H]$ ?

(2) Do $\ord(a)$ and $\gcd_H[G:H]$ always have the same prime factors?

Answer 1

The answer to Q1 is yes, the order of $a$ might be smaller than the gcd: Let $G=\langle x,y\mid x^8=y^2=1,x^y=x^3\rangle$ be the semidihedral group of order $16$. Let $a=[x]\in G_{\text{ab}}=C_2\times C_2$. Clearly $\lvert a\rvert=2$. Let $X$ be the set of subgroups $H$ with $V_H(a)=0$. I claim that $\gcd_{H\in X}([G:H])\geq 4$: (Clearly $1\in X$ so the gcd is welldefined). One just have to see that $X$ does not contain any of the $3$ maximal subgroups $H_1=\langle x\rangle$, $H_2=\langle x^2,y\rangle$ and $H_3=\langle x^2,xy\rangle$ of $G$. If I did the transfer computation correctly one has $V_1(a)=[x^4]\neq 0$, $V_2(a)=[x^2]\neq 0$, $V_3(a)=[x^2]\neq 0$ proving the claim. (One can actually skip $H_3$ here since there is an automorphism of $G$ which swaps $H_2$ and $H_3$.)

Update The answer to Q2 is yes if $G$ is finite (I leave the profinite case to the interested reader....): Let $a\in G^{\text{ab}}$ and $X$ be the set of subgroups of $G$ such that $V_H(a)=0$. Note that $1\in X$, so $\lvert G\rvert a=0$. Assume that $p$ is a prime which does not divide $\text{ord}(a)$, and let $S$ be a Sylow $p$-subgroup of $G$. Then $p$ does not divide $\text{ord}(V_S(a))$. On the other hand $V_S(a)\in S^{\text{ab}}$ has $p$-power order so $V_S(a)=0$. Hence $S\in X$ so $p$ does not divide $\gcd_{H\in X} [G:H]$.