Consider the following group homomorphism $\rho$, where $R$ is an abelian group,

\begin{align*} \rho:&R\rightarrow R^n\\ \rho(r)=&(2r,2r,\cdots,2r). \end{align*} Show that $R^n/Im(\rho)=R^{n-1}\bigoplus R/2R$.

I'm confused if it's an equality or if I should show that $R^n/Im(\rho)$ and $R^{n-1}\bigoplus R/2R$ are isomorphic. Also, in case it was an isomorphism, I was thinking of using the first isomorphism theorem and defining an homomorphism that has $Im(\rho)$ as its kernel, but I can't think of anyone like that.

Any other hint would be very appreciated. Thanks!


Most likely you are to show an isomorphism. An intuitive idea is to first think of the case when $R = \mathbb{Z}$. Now note that in this case we are working with free $\mathbb{Z}$-modules, and in particular we have a change of basis $f$ of $R^n$ via $e_1 \mapsto e_1 + \dots + e_n =: v$ and $e_i \mapsto e_i$ for $i> 1$. Thus $im \rho = 2v\mathbb{Z}$ and:

$$ \mathbb{Z}^n/2v\mathbb{Z} \stackrel{(via f^{-1})}{\simeq} \mathbb{Z}^n/2e_1\mathbb{Z} = \frac{\mathbb{Z} \oplus \mathbb{Z}\oplus \dots \oplus \mathbb{Z}}{2\mathbb{Z} \oplus 0 \oplus \dots \oplus 0} = \frac{\mathbb{Z}}{2\mathbb{Z}} \oplus \frac{\mathbb{Z}}{0} \oplus \dots \oplus \frac{\mathbb{Z}}{0} = \mathbb{Z}/(2) \oplus \mathbb{Z}^{n-1} $$

With the same ideas, let $g : R^n \to R^n$ defined as

$$ \begin{align} g(r_1, \dots, r_n) := (r_1, r_2 + r_1, \dots, r_n + r_1). \end{align} $$

be an automorphism of $R^n$. Note that when $R$ is $\mathbb{Z}$, it coincides with the function $f$ we previously defined. Now,

$$ R/2R \oplus R^{n-1} \simeq \frac{R}{2R} \oplus \frac{R}{0} \oplus \dots \oplus \frac{R}{0} \simeq \frac{R \oplus R \oplus \dots \oplus R}{2R \oplus 0 \oplus \dots \oplus 0} = R^n/S $$

with $S = \{(2r, 0, \dots, 0) : r \in R\}$. It suffices to see, then, that $im \rho = g(S)$. In effect,

$$ g(2r,0, \dots, 0) = (2r, \dots, 2r) $$

for all $r \in R$.


By trial and error, take the morphism $\pi:R^n\to R^{n-1}\bigoplus R/2R$ given by $(r_1,\ldots,r_n)\mapsto(r_1-r_2, r_2-r_3,\ldots, r_{n-1}-r_{n}, \bar{r}_n)$, where $\bar{r}_n$ is the class of $r_n$. You can see that $\text{Im}(\rho)=\ker\pi$. The map $\pi$ is surjective, you can solve the system $r_1-r_2=s_1,\ldots, r_{n-1}-r_{n}=s_{n-1}, \bar{r}_n=s_n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.