Let $G$ be a finite abelian group. By the structural theorem of finitely-generated abelian groups, we know that $G$ has canonical decomposition $$G\cong \mathbf{Z}_{d_1}\oplus\cdots\oplus\mathbf{Z}_{d_n}$$ where $d_1\mid \dots\mid d_n$. Define $c(G):=n$ to be the number of constituents of $G$.

Now suppose $\chi:G\to S^1$ is a group character (i.e. a group homomorphism), then $\ker(\chi)$ is a subgroup of $G$, how to show that $c(\ker\chi)\geq c(G)-1$?

  • $\begingroup$ I'm interested in where this problem comes from. There seems to be a more general structure behind it. May you provide a reference? $\endgroup$ – Cave Johnson Sep 23 '16 at 2:10
  • $\begingroup$ @CaveJohnson It comes from this paper: sciencedirect.com/science/article/pii/0097316595900241. In fact, it follows from a more general theorem in this paper. $\endgroup$ – Xiang Yu Sep 23 '16 at 2:14
  • $\begingroup$ I see. The paper stated that $c(\ker\varphi)\geq c(G)-c(H)$, which is exactly the lemma I established in my answer :) But I have the feeling that some more interesting connections between $c(G)$ and $c(H)$ is in the dark. $\endgroup$ – Cave Johnson Sep 23 '16 at 2:20

The function $c$ may be intuitively thought of as the "dimension", in sense of the following

Lemma. Let $G$ be a finite abelian $p$-group. Then for any $H\lhd G$ we have $$c(G/H)\geq c(G)-c(H)$$
Proof. Let $k=c(G/H)$, and suppose $G/H\cong\bigoplus\limits_{i=1}^k\mathbb{Z}_{p^{\alpha_{~i}}}$. So we may say every element in $G/H$ can be expressed as $\sum\limits_{i=1}^k n_i(g_i+H)=\sum\limits_{i=1}^k n_i g_i+H$, where $n_i\in\mathbb{Z}$. This means that every element in $G$ can be expressed as the "linear" combination of $k+c(H)$ elements. So $G$ is a homomorphic image of $\mathbb{Z}^{k+c(H)}$, hence cannot have more than $k+c(H)$ components in the canonical decomposition. Consequently we get $c(G)\leq k+c(H)$, which is to be proved.

Now note that in the decomposition $$G\cong\bigoplus_{i\leq n,~j\leq k}\mathbb{Z}_{p_i^{\alpha_{~ij}}}$$ we have $c(G)=k$, where $0\leq\alpha_{i1}\leq\alpha_{i2}\leq\cdots\leq\alpha_{ik}$ , $\alpha_{ik}\not=0$, and at least one of $\alpha_{i1}\not=0$. From this point of view we see that it suffices to prove the statement for finite abelian $p$-groups. WLOG we may assume $$G\cong\bigoplus_{j=1}^k\mathbb{Z}_{p^{\alpha_{~j}}}$$Suppose the generator of these direct summands is $g_{j}$. By choosing generators properly, we may safely say that $\chi$ sends $g_{j}$ to $$\exp({2\pi \sqrt{-1}/p^{\beta_{j}}})$$ Here $\beta_j\leq\alpha_j$. Now we see $G/\ker\chi\cong\text{im }\chi\cong\mathbb{Z}_{p^\beta}$, where $\beta=\max\beta_j$. So from the Lemma we get $c(\ker\chi)\geq c(G)-c(G/\ker\chi)=c(G)-1$.


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