# The condition of determining a group homomorphism between two finite abelian group

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

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.

• 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. – bfhaha Jan 18 '17 at 18:42
• 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 $G\to G/K'$. – Robert Chamberlain Jan 18 '17 at 19:14
• @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 $\psi_0 = \psi\circ\rho$). – GaryMak Jan 17 '18 at 13:49