There is an exercise (Exercise 10.25 in Gallian's Contemporary Abstract Algebra 8/e) which asks: How many group homomorphisms are there from $\Bbb{Z}_{20}$ onto $\Bbb{Z}_{10}$? How many are there to $\Bbb{Z}_{10}$?

The answer is $\varphi(10)=4$ and $10$, respectively. It is sufficient to determinet the image of the generator of $\Bbb{Z}_{20}$. But I found an example which seems to be a homomorphism, but it is not. The example is: $\theta:\Bbb{Z}_3\times \Bbb{Z}_2\to S_3$, $\theta(1,0)=(123)$ and $\theta(0,1)=(12)$. This mapping preserve the operation. But it is not well-defined. $$(123)(12) =\theta(1,0)\cdot \theta(0,1) =\theta((1,0)+(0,1)) =\theta(1,1) =\theta((0,1)+(1,0)) =\theta(0,1)\cdot \theta(1,0) =(12)\cdot (123).$$

Is the following statement true? Let $G\cong \Bbb{Z}_{p_1^{r_1}}\times \Bbb{Z}_{p_2^{r_2}}\times \cdots \times \Bbb{Z}_{p_s^{r_s}}$ and $H$ be two finite additive abelian groups. If a mapping $f:G\to H$ satisfy $p_1^{r_1}f(1,0,...,0)=0_{H}$, $p_2^{r_2}f(0,1,0,...,0)=0_{H}$, ..., $p_s^{r_s}f(0,0,..., 0,1)=0_{H}$, then $f$ must be a homomorphism from $G$ to $H$.

Remark. There is an error in my question. I can't just only define $f$ as a ``function''. For example, define $f:\Bbb{Z}_2\to \Bbb{Z}_2$ by $f(0)=f(1)=1$. Which satisfy the condition. But it is not a homomorphism.


1 Answer 1


The statement is true and you could probably spend hours trying to prove it, but it's much nicer to use group presentations which give us something much more general (apologies, I will be using multiplicative notation).

You may be aware that any group $G$ can be written $\langle P|R\rangle =F(P)/\langle\langle R\rangle\rangle$ where $P\subset G$ generates $G$, $F(P)$ is the free group on $P$, and $\langle\langle R\rangle\rangle$ is the smallest normal group in $F(P)$ containing the set $R\subseteq F(P)$. $P$ is a set of generators of $G$ and $R$ is a set of relators.

In general an arbitrary map $\psi_0:P\to H$ defines a homomorphism $\psi:G\to H$ if and only if $\psi_0(r)=1_H$ for each $r\in R$.

In your question specifically $G$ is generated by $e_1,\ldots,e_s$, where $e_i$ can be thought of as an $s$-tuple with $1$ in the $i^{th}$ entry and $0$ elsewhere. One set of relators that define $G$ is $R=\{e_ie_je_i^{-1}e_j^{-1},e_i^{p_i^{r_i}}|1\le i,j\le s\}$. The $e_ie_je_i^{-1}e_j^{-1}$ just mean that $G$ is abelian and since $H$ is abelian, these certainly map under $f$ to $1_H$. The $e_i^{p_i^{r_i}}$ just mean that $e_i$ has order $p_i^{r_i}$ and by assumption, these map under $f$ to $1_H$: so $f$ does indeed define a homomorphism.

  • $\begingroup$ Thanks. Could you please tell me where did you read the lemma you use ($\psi:G\to H\Leftrightarrow \psi(r)=1_H$). It seems to be the universal property. But there is a little bit difference. $\endgroup$
    – bfhaha
    Commented Jan 18, 2017 at 18:42
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    $\begingroup$ It's actually not very hard to prove and is usually done in any introductory book / course on group presentations. $r=1_G$ in $G$ so $\psi(r)=1_H$ is immediate. For the other direction, by definition of free group $\psi_0$ defines a homomorphism $\psi_1:F(P)\to H$. The kernel $K$ of this map contains each $r\in R$ so contains $\langle\langle R\rangle\rangle$. Therefore $H\cong F(P)/K\cong (F(P)/\langle\langle R\rangle\rangle)/(K/\langle\langle R\rangle\rangle)\cong G/K'$ so $\psi$ can be identified as the natural map $G\to G/K'$. $\endgroup$ Commented Jan 18, 2017 at 19:14
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    $\begingroup$ @RobertChamberlain This is a great answer; I would just suggest that it might be helpful in the first "direction" of the proof in the comment above, to write something like: under the presentation hom, say $\rho:F(P)\longrightarrow G$, every $r$ in $R$ is mapped to $1_G$ and so $\psi_0(r) = 1_H$ (since $\psi_0 = \psi\circ\rho$). $\endgroup$
    – GaryMak
    Commented Jan 17, 2018 at 13:49

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