Show that $R^n/Im(rho)=R^{n-1}bigoplus R/2R$, where $R$ is an abelian group and $rho$ is the following...
$begingroup$
Consider the following group homomorphism $rho$, where $R$ is an abelian group,
begin{align*}
rho:&Rrightarrow 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!
group-theory abelian-groups group-isomorphism group-homomorphism direct-sum
edited Dec 11 '18 at 0:46
amWhy
1
asked Dec 11 '18 at 0:41
user392559user392559
37618
$endgroup$
add a comment |
$begingroup$
Consider the following group homomorphism $rho$, where $R$ is an abelian group,
begin{align*}
rho:&Rrightarrow 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!
group-theory abelian-groups group-isomorphism group-homomorphism direct-sum
edited Dec 11 '18 at 0:46
amWhy
1
asked Dec 11 '18 at 0:41
user392559user392559
37618
$endgroup$
add a comment |
$begingroup$
Consider the following group homomorphism $rho$, where $R$ is an abelian group,
begin{align*}
rho:&Rrightarrow 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!
group-theory abelian-groups group-isomorphism group-homomorphism direct-sum
edited Dec 11 '18 at 0:46
amWhy
1
asked Dec 11 '18 at 0:41
user392559user392559
37618
$endgroup$
Consider the following group homomorphism $rho$, where $R$ is an abelian group,
begin{align*}
rho:&Rrightarrow 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!
group-theory abelian-groups group-isomorphism group-homomorphism direct-sum
group-theory abelian-groups group-isomorphism group-homomorphism direct-sum
edited Dec 11 '18 at 0:46
amWhy
1
asked Dec 11 '18 at 0:41
user392559user392559
37618
edited Dec 11 '18 at 0:46
amWhy
1
asked Dec 11 '18 at 0:41
user392559user392559
37618
edited Dec 11 '18 at 0:46
amWhy
1
edited Dec 11 '18 at 0:46
amWhy
1
edited Dec 11 '18 at 0:46
amWhy
1
1
asked Dec 11 '18 at 0:41
user392559user392559
37618
asked Dec 11 '18 at 0:41
user392559user392559
37618
asked Dec 11 '18 at 0:41
user392559user392559
37618
37618
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $im rho = 2vmathbb{Z}$ and:
$$
mathbb{Z}^n/2vmathbb{Z} stackrel{(via f^{-1})}{simeq} mathbb{Z}^n/2e_1mathbb{Z} = frac{mathbb{Z} oplus mathbb{Z}oplus dots oplus mathbb{Z}}{2mathbb{Z} oplus 0 oplus dots oplus 0} = frac{mathbb{Z}}{2mathbb{Z}} oplus frac{mathbb{Z}}{0} oplus dots oplus frac{mathbb{Z}}{0} = mathbb{Z}/(2) oplus mathbb{Z}^{n-1}
$$
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 $r in R$.
answered Dec 11 '18 at 1:36
Guido A.Guido A.
7,4001730
$endgroup$
add a comment |
$begingroup$
By trial and error, take the morphism $pi:R^nto R^{n-1}bigoplus R/2R$ 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 $text{Im}(rho)=kerpi$. The map $pi$ is surjective, you can solve the system $r_1-r_2=s_1,ldots, r_{n-1}-r_{n}=s_{n-1}, bar{r}_n=s_n$.
answered Dec 11 '18 at 1:36
user90189user90189
752616
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add a comment |
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2 Answers
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2 Answers
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active
oldest
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votes
$begingroup$
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 $im rho = 2vmathbb{Z}$ and:
$$
mathbb{Z}^n/2vmathbb{Z} stackrel{(via f^{-1})}{simeq} mathbb{Z}^n/2e_1mathbb{Z} = frac{mathbb{Z} oplus mathbb{Z}oplus dots oplus mathbb{Z}}{2mathbb{Z} oplus 0 oplus dots oplus 0} = frac{mathbb{Z}}{2mathbb{Z}} oplus frac{mathbb{Z}}{0} oplus dots oplus frac{mathbb{Z}}{0} = mathbb{Z}/(2) oplus mathbb{Z}^{n-1}
$$
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 $r in R$.
answered Dec 11 '18 at 1:36
Guido A.Guido A.
7,4001730
$endgroup$
add a comment |
$begingroup$
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 $im rho = 2vmathbb{Z}$ and:
$$
mathbb{Z}^n/2vmathbb{Z} stackrel{(via f^{-1})}{simeq} mathbb{Z}^n/2e_1mathbb{Z} = frac{mathbb{Z} oplus mathbb{Z}oplus dots oplus mathbb{Z}}{2mathbb{Z} oplus 0 oplus dots oplus 0} = frac{mathbb{Z}}{2mathbb{Z}} oplus frac{mathbb{Z}}{0} oplus dots oplus frac{mathbb{Z}}{0} = mathbb{Z}/(2) oplus mathbb{Z}^{n-1}
$$
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 $r in R$.
answered Dec 11 '18 at 1:36
Guido A.Guido A.
7,4001730
$endgroup$
add a comment |
$begingroup$
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 $im rho = 2vmathbb{Z}$ and:
$$
mathbb{Z}^n/2vmathbb{Z} stackrel{(via f^{-1})}{simeq} mathbb{Z}^n/2e_1mathbb{Z} = frac{mathbb{Z} oplus mathbb{Z}oplus dots oplus mathbb{Z}}{2mathbb{Z} oplus 0 oplus dots oplus 0} = frac{mathbb{Z}}{2mathbb{Z}} oplus frac{mathbb{Z}}{0} oplus dots oplus frac{mathbb{Z}}{0} = mathbb{Z}/(2) oplus mathbb{Z}^{n-1}
$$
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 $r in R$.
answered Dec 11 '18 at 1:36
Guido A.Guido A.
7,4001730
$endgroup$
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 $im rho = 2vmathbb{Z}$ and:
$$
mathbb{Z}^n/2vmathbb{Z} stackrel{(via f^{-1})}{simeq} mathbb{Z}^n/2e_1mathbb{Z} = frac{mathbb{Z} oplus mathbb{Z}oplus dots oplus mathbb{Z}}{2mathbb{Z} oplus 0 oplus dots oplus 0} = frac{mathbb{Z}}{2mathbb{Z}} oplus frac{mathbb{Z}}{0} oplus dots oplus frac{mathbb{Z}}{0} = mathbb{Z}/(2) oplus mathbb{Z}^{n-1}
$$
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 $r in R$.
answered Dec 11 '18 at 1:36
Guido A.Guido A.
7,4001730
edited Dec 11 '18 at 1:48
answered Dec 11 '18 at 1:36
Guido A.Guido A.
7,4001730
answered Dec 11 '18 at 1:36
Guido A.Guido A.
7,4001730
answered Dec 11 '18 at 1:36
Guido A.Guido A.
7,4001730
7,4001730
add a comment |
add a comment |
$begingroup$
By trial and error, take the morphism $pi:R^nto R^{n-1}bigoplus R/2R$ 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 $text{Im}(rho)=kerpi$. The map $pi$ is surjective, you can solve the system $r_1-r_2=s_1,ldots, r_{n-1}-r_{n}=s_{n-1}, bar{r}_n=s_n$.
answered Dec 11 '18 at 1:36
user90189user90189
752616
$endgroup$
add a comment |
$begingroup$
By trial and error, take the morphism $pi:R^nto R^{n-1}bigoplus R/2R$ 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 $text{Im}(rho)=kerpi$. The map $pi$ is surjective, you can solve the system $r_1-r_2=s_1,ldots, r_{n-1}-r_{n}=s_{n-1}, bar{r}_n=s_n$.
answered Dec 11 '18 at 1:36
user90189user90189
752616
$endgroup$
add a comment |
$begingroup$
By trial and error, take the morphism $pi:R^nto R^{n-1}bigoplus R/2R$ 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 $text{Im}(rho)=kerpi$. The map $pi$ is surjective, you can solve the system $r_1-r_2=s_1,ldots, r_{n-1}-r_{n}=s_{n-1}, bar{r}_n=s_n$.
answered Dec 11 '18 at 1:36
user90189user90189
752616
$endgroup$
By trial and error, take the morphism $pi:R^nto R^{n-1}bigoplus R/2R$ 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 $text{Im}(rho)=kerpi$. The map $pi$ is surjective, you can solve the system $r_1-r_2=s_1,ldots, r_{n-1}-r_{n}=s_{n-1}, bar{r}_n=s_n$.
answered Dec 11 '18 at 1:36
user90189user90189
752616
edited Dec 11 '18 at 1:47
answered Dec 11 '18 at 1:36
user90189user90189
752616
answered Dec 11 '18 at 1:36
user90189user90189
752616
answered Dec 11 '18 at 1:36
user90189user90189
752616
752616
add a comment |
add a comment |
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