Intersection of subgroups of a finite abelian group Let $G$ be a finite abelian group and let $g \in G$ be a non-trivial element. I want to show that the intersection over all subgroups $G'$ of $G$ such that $G/G'$ is cyclic  and $g \not \in G'$  is trivial.
My attempt: By the Fundamental theorem of abelian groups, $G \cong \mathbb Z/n_1 \mathbb Z \times \cdots \times  \mathbb Z/n_m \mathbb Z$. We prove by induction on the number of cyclic factors of $G$.
For the base case ($m=1$), $G$ is cyclic. Then if we let $G'=\{e\}$,  $G/G'$ is cyclic and $g \not \in G'$. Thus, the proposition is true in this case.
Now for the general case, write  $G = \mathbb Z/n_1 \mathbb Z \times \cdots \times  \mathbb Z/n_m \mathbb Z$ as before and let $g=(g_1, \ldots, g_m)$. Since $g$ is non-trivial, at least of its components is not zero. WLOG, assume $g_1 \neq 0$. Then $h:=(g_1, \ldots, g_{m-1})$ is a non-zero element of $H:=\mathbb Z/n_1 \mathbb Z \times \cdots \times  \mathbb Z/n_{m-1} \mathbb Z$. By the inudction hypothesis, the intersection over all subgroups $H'$ of $H$ such that $H/H'$ is cyclic  and $h \not \in H'$  is trivial.
I am not sure how to proceed from here and I will be very grateful for any advice on this attempt or some ideas for an alternative attempt.
 A: I think the key is the $k=2$ case. Once you figure out how to do the $n=2$ case, the general case follows the same idea. So let me do the $k=2$ case explicitly.
Let $G=C_{n_1}\times C_{n_2}$, with $n_1|n_2$. Let $x_1$ and $x_2$ be the corresponding generators, and let $g=(g_1,g_2)$ be a nontrivial element of $G$.
If $g_1\neq e$ and $g_2\neq e$, then we can take $H_1=C_{n_1}\times\{e\}$ and $H_2=\{e\}\times C_{n_2}$; each of those excludes $g$, $G/H_i$ is cyclic, and $H_1\cap H_2=\{e\}$, so we are fine.
If either $g_1$ or $g_2$ are trivial, let $y=(x_1,x_2^{n_2/n_1})$, and let $H=\langle y \rangle$. Note that $H\cap H_1=H\cap H_2 = \{e\}$, because both components have order $n_1$. This also means that $g\notin H$. Finally, $(e,x_2)H$ has order $n_2$ in $G/H$ (because $H$ does not contain any elements of $\{e\}\times C_{n_2}$ that are nontrivial), and $G/H$ has order $n_2$, so $G/H$ is cyclic. Now, either $H_1$ or $H_2$ do not contain $g$, so whichever one does not contain $g$, together with $H$, show the intersection you want is trivial. This finishes the $k=2$ case.
Note: To better mimic the general case, we could do the following instead: if $g_2$ is trivial, we replace $x_1$ and $x_2$ with $y$ and $x_2$ to get new "coordinates" for $G$: $G\cong H\times H_2$, but in these coordinates, both components of $g$ are nontrivial, so we are reduced to the first case. And if $g_1$ is trivial, then we take $H_1$ and $H$ as our subgroups.

So, now let $G=C_{n_1}\times\cdots\times C_{n_k}\times C_{n_{k+1}}$ with $n_1|\cdots|n_k|n_{k+1}$, $x_i$ a fixed generator for $C_{n_i}$, and let $g=(g_1,\ldots,g_{k+1})\neq e_G$. Let $H=C_{n_1}\times\cdots\times C_{n_k}$, so $G=H\times C_{n_{k+1}}$, and write $u=(g_1,\ldots,g_k)$. We identify $g$ with $(u,g_{k+1})$ and $H$ with the corresponding subgroup of $G$ when necessary (I'm going to play fast and loose with the internal vs. external direct product, but being more careful is going to be annoying; I hope that's okay).  We assume by induction that if $u\neq e_H$, then we can find subgroups of $H$ that do not include $u$, with cyclic quotient, with trivial intersection.
If $u\neq e$ and $g_{k+1}\neq e$, then let $K_1,\ldots,K_r$ be subgroups of $H$ with $u\notin K_i$, $H/K_i$ cyclic, and $K_1\cap\cdots\cap K_r$ trivial. Then we have that $K_i\times C_{n_{k+1}}$ does not contain $g$, and $G/(K_i\times C_{n_{k+1}})\cong H/K_i$ is cyclic. In addition, $g\notin H\times\{e\}$, and $G/(H\times\{e\})\cong C_{n_{k+1}}$ is cyclic. Now we notice that $H\times\{e\}\cap (K_1\times C_{n_{k+1}})\cap\cdots\cap (K_r\times C_{n_{k+1}})$ is trivial, so we are done.
If either $u=e$ or $g_{k+1}=e$, then for $i=1,\ldots,k$, let $y_i = x_{i}x_{n_{k+1}}^{n_{k+1}/n_i}$, and let $H_i = \langle x_1,\ldots,x_{i-1},y_i,x_{i+1},\ldots,x_k\rangle$.
Note that an element of $H_i$, viewed as a $(k+1)$-tuple, has trivial $i$th coordinate if and only if it has trivial $(k+1)$st coordinate. In particular, $H\cap H_1\cap\cdots\cap H_k$ is trivial. And $G/H_i$ has order $n_{k+1}$, and as in the $k=2$ case, the element $x_{k+1}H_{i}$ has order $n_{k+1}$, so $G/H_i$ is cyclic.
If $u\neq e$ and $g_{k+1}=e$, then say the $i$th component of $u$ is nontrivial. Then replacing $H$ with $H_i$ we have that $G\cong H_i\times\langle x_{k+1}\rangle$, but now expressing $g$ in terms of $x_1,\ldots,x_{i-1},y_i,x_{i+1},\ldots,x_{k+1}$ will have both the $H_i$ and the $x_{k+1}$ component nontrivial, so we are reduced to the first case and we are done.
And if $u=e$ and $g_{k+1}\neq e$, then each of $H$, $H_1,\ldots,H_k$ are subgroups that do not contain $g$, with $G/H$ and $G/H_j$ cyclic, and $H\cap H_1\cap\cdots\cap H_k$ trivial, so we are done.
