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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!

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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 $r \in R$.

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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 $r_1-r_2=s_1,\ldots, r_{n-1}-r_{n}=s_{n-1}, \bar{r}_n=s_n$.

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