Mixing
Explain why $U_{44} cong (mathbb{Z}_{10} oplus mathbb{Z}_2) $.
$begingroup$
Explain why $newcommand{Z}{mathbb{Z}} U_{44} cong (Z_{10} oplus Z_2) $
I know that $Z_{20} cong (Z_{10} oplus Z_2)$, so if I can show $U_{44} cong Z_{20}$, then I can conclude that $U_{44} cong (Z_{10} oplus Z_2) $ since isomorphism is transitive.
In order to show $U_{44} cong Z_{20}$, I need to show that the function $ varphi: U_{44} to Z_{20}$ is bijective and structure preserved. I know that $U_{44} text{ and } Z_{20}$ have order $20$, so the function can be bijective, but it's not enough to show $varphi: U_{44} to Z_{20} $ is bijective.
I wonder if anyone can show me how to show $ varphi: U_{44} to Z_{20} $ is bijective and structure preserving.
abstract-algebra group-theory finite-groups group-isomorphism
edited Jan 28 at 15:03
Martin Sleziak
44.9k10122276
asked Apr 17 '13 at 23:58
Diane VanderwaifDiane Vanderwaif
1,01032053
$endgroup$
add a comment |
$begingroup$
Explain why $newcommand{Z}{mathbb{Z}} U_{44} cong (Z_{10} oplus Z_2) $
I know that $Z_{20} cong (Z_{10} oplus Z_2)$, so if I can show $U_{44} cong Z_{20}$, then I can conclude that $U_{44} cong (Z_{10} oplus Z_2) $ since isomorphism is transitive.
In order to show $U_{44} cong Z_{20}$, I need to show that the function $ varphi: U_{44} to Z_{20}$ is bijective and structure preserved. I know that $U_{44} text{ and } Z_{20}$ have order $20$, so the function can be bijective, but it's not enough to show $varphi: U_{44} to Z_{20} $ is bijective.
I wonder if anyone can show me how to show $ varphi: U_{44} to Z_{20} $ is bijective and structure preserving.
abstract-algebra group-theory finite-groups group-isomorphism
edited Jan 28 at 15:03
Martin Sleziak
44.9k10122276
asked Apr 17 '13 at 23:58
Diane VanderwaifDiane Vanderwaif
1,01032053
$endgroup$
2
$begingroup$
Actually, begin{align*} Bbb{Z}/20Bbb{Z}&congBbb{Z}/4Bbb{Z}timesBbb{Z}/5Bbb{Z}\ ¬congBbb{Z}/10Bbb{Z}timesBbb{Z}/2Bbb{Z}. end{align*} ($gcd(10,2) = 2neq 1$)
$endgroup$
– Stahl
Apr 18 '13 at 0:22
add a comment |
$begingroup$
Explain why $newcommand{Z}{mathbb{Z}} U_{44} cong (Z_{10} oplus Z_2) $
I know that $Z_{20} cong (Z_{10} oplus Z_2)$, so if I can show $U_{44} cong Z_{20}$, then I can conclude that $U_{44} cong (Z_{10} oplus Z_2) $ since isomorphism is transitive.
In order to show $U_{44} cong Z_{20}$, I need to show that the function $ varphi: U_{44} to Z_{20}$ is bijective and structure preserved. I know that $U_{44} text{ and } Z_{20}$ have order $20$, so the function can be bijective, but it's not enough to show $varphi: U_{44} to Z_{20} $ is bijective.
I wonder if anyone can show me how to show $ varphi: U_{44} to Z_{20} $ is bijective and structure preserving.
abstract-algebra group-theory finite-groups group-isomorphism
edited Jan 28 at 15:03
Martin Sleziak
44.9k10122276
asked Apr 17 '13 at 23:58
Diane VanderwaifDiane Vanderwaif
1,01032053
$endgroup$
Explain why $newcommand{Z}{mathbb{Z}} U_{44} cong (Z_{10} oplus Z_2) $
I know that $Z_{20} cong (Z_{10} oplus Z_2)$, so if I can show $U_{44} cong Z_{20}$, then I can conclude that $U_{44} cong (Z_{10} oplus Z_2) $ since isomorphism is transitive.
In order to show $U_{44} cong Z_{20}$, I need to show that the function $ varphi: U_{44} to Z_{20}$ is bijective and structure preserved. I know that $U_{44} text{ and } Z_{20}$ have order $20$, so the function can be bijective, but it's not enough to show $varphi: U_{44} to Z_{20} $ is bijective.
I wonder if anyone can show me how to show $ varphi: U_{44} to Z_{20} $ is bijective and structure preserving.
abstract-algebra group-theory finite-groups group-isomorphism
abstract-algebra group-theory finite-groups group-isomorphism
edited Jan 28 at 15:03
Martin Sleziak
44.9k10122276
asked Apr 17 '13 at 23:58
Diane VanderwaifDiane Vanderwaif
1,01032053
edited Jan 28 at 15:03
Martin Sleziak
44.9k10122276
asked Apr 17 '13 at 23:58
Diane VanderwaifDiane Vanderwaif
1,01032053
edited Jan 28 at 15:03
Martin Sleziak
44.9k10122276
edited Jan 28 at 15:03
Martin Sleziak
44.9k10122276
edited Jan 28 at 15:03
Martin Sleziak
44.9k10122276
44.9k10122276
asked Apr 17 '13 at 23:58
Diane VanderwaifDiane Vanderwaif
1,01032053
asked Apr 17 '13 at 23:58
Diane VanderwaifDiane Vanderwaif
1,01032053
asked Apr 17 '13 at 23:58
Diane VanderwaifDiane Vanderwaif
1,01032053
1,01032053
2
$begingroup$
Actually, begin{align*} Bbb{Z}/20Bbb{Z}&congBbb{Z}/4Bbb{Z}timesBbb{Z}/5Bbb{Z}\ ¬congBbb{Z}/10Bbb{Z}timesBbb{Z}/2Bbb{Z}. end{align*} ($gcd(10,2) = 2neq 1$)
$endgroup$
– Stahl
Apr 18 '13 at 0:22
add a comment |
2
$begingroup$
Actually, begin{align*} Bbb{Z}/20Bbb{Z}&congBbb{Z}/4Bbb{Z}timesBbb{Z}/5Bbb{Z}\ ¬congBbb{Z}/10Bbb{Z}timesBbb{Z}/2Bbb{Z}. end{align*} ($gcd(10,2) = 2neq 1$)
$endgroup$
– Stahl
Apr 18 '13 at 0:22
2
2
$begingroup$
Actually, begin{align*} Bbb{Z}/20Bbb{Z}&congBbb{Z}/4Bbb{Z}timesBbb{Z}/5Bbb{Z}\ ¬congBbb{Z}/10Bbb{Z}timesBbb{Z}/2Bbb{Z}. end{align*} ($gcd(10,2) = 2neq 1$)
$endgroup$
– Stahl
Apr 18 '13 at 0:22
$begingroup$
Actually, begin{align*} Bbb{Z}/20Bbb{Z}&congBbb{Z}/4Bbb{Z}timesBbb{Z}/5Bbb{Z}\ ¬congBbb{Z}/10Bbb{Z}timesBbb{Z}/2Bbb{Z}. end{align*} ($gcd(10,2) = 2neq 1$)
$endgroup$
– Stahl
Apr 18 '13 at 0:22
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Be careful: $$quad mathbb Z_{10} times mathbb Z_2 notcong mathbb Z_{20}; quadtext{since};gcd(2, 10) = 2neq 1$$
Use the facts that
- $U(44)= U(4cdot 11)cong U(4)times U(11)$, since $4$ and $11$ are relatively prime.
- $U(4)cong mathbb Z_2$ and $U(11)cong mathbb Z_{10}$.
To conclude that $$U(44) cong mathbb Z_{2}times mathbb Z_{10}$$
answered Apr 18 '13 at 0:13
NamasteNamaste
1
$endgroup$
$begingroup$
In my answer, $times$ denotes what you might know as the external direct product: $otimes$ or in the case of $mathbb Z_n times mathbb Z_m$ it denotes what some call the external direct sum: i.e., $mathbb Z_{10} otimes mathbb Z_2 notcong mathbb Z_{20},$ etc...with the conclusion that $U_{44} cong mathbb Z_2 otimes mathbb Z_{10}$. I apologize for any confusion notation may be causing!
$endgroup$
– Namaste
Apr 18 '13 at 1:39
$begingroup$
don't worry, I understand it perfectly, thank you very much for your answer.
$endgroup$
– Diane Vanderwaif
Apr 18 '13 at 2:09
add a comment |
$begingroup$
Hint: List the elements and compute their order. From this it should be quite clear.
answered Apr 18 '13 at 0:03
IslandsIslands
457213
$endgroup$
add a comment |
$begingroup$
In general, for $n=p_1^{e_1}cdots p_k^{e_k}$, $U_ncong(mathbb{Z}/p_1^{e_1}mathbb{Z})^timestimescdotstimes(mathbb{Z}/p_k^{e_k}mathbb{Z})^times$ by the Chinese Remainder Theorem.
answered Apr 18 '13 at 0:07
Warren MooreWarren Moore
3,2731128
$endgroup$
add a comment |
$begingroup$
Note that $U(44)= U(11*4)cong {mathbb Z_{10}} oplus {mathbb Z_2}$, since $gcd(11,4)=1.$
In general, if $gcd(a,b)=1$, then $U(ab)cong {mathbb Z_a}oplus{mathbb Z_b}$
answered Apr 18 '13 at 0:04
lukeluke
969819
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Be careful: $$quad mathbb Z_{10} times mathbb Z_2 notcong mathbb Z_{20}; quadtext{since};gcd(2, 10) = 2neq 1$$
Use the facts that
- $U(44)= U(4cdot 11)cong U(4)times U(11)$, since $4$ and $11$ are relatively prime.
- $U(4)cong mathbb Z_2$ and $U(11)cong mathbb Z_{10}$.
To conclude that $$U(44) cong mathbb Z_{2}times mathbb Z_{10}$$
answered Apr 18 '13 at 0:13
NamasteNamaste
1
$endgroup$
$begingroup$
In my answer, $times$ denotes what you might know as the external direct product: $otimes$ or in the case of $mathbb Z_n times mathbb Z_m$ it denotes what some call the external direct sum: i.e., $mathbb Z_{10} otimes mathbb Z_2 notcong mathbb Z_{20},$ etc...with the conclusion that $U_{44} cong mathbb Z_2 otimes mathbb Z_{10}$. I apologize for any confusion notation may be causing!
$endgroup$
– Namaste
Apr 18 '13 at 1:39
$begingroup$
don't worry, I understand it perfectly, thank you very much for your answer.
$endgroup$
– Diane Vanderwaif
Apr 18 '13 at 2:09
add a comment |
$begingroup$
Be careful: $$quad mathbb Z_{10} times mathbb Z_2 notcong mathbb Z_{20}; quadtext{since};gcd(2, 10) = 2neq 1$$
Use the facts that
- $U(44)= U(4cdot 11)cong U(4)times U(11)$, since $4$ and $11$ are relatively prime.
- $U(4)cong mathbb Z_2$ and $U(11)cong mathbb Z_{10}$.
To conclude that $$U(44) cong mathbb Z_{2}times mathbb Z_{10}$$
answered Apr 18 '13 at 0:13
NamasteNamaste
1
$endgroup$
$begingroup$
In my answer, $times$ denotes what you might know as the external direct product: $otimes$ or in the case of $mathbb Z_n times mathbb Z_m$ it denotes what some call the external direct sum: i.e., $mathbb Z_{10} otimes mathbb Z_2 notcong mathbb Z_{20},$ etc...with the conclusion that $U_{44} cong mathbb Z_2 otimes mathbb Z_{10}$. I apologize for any confusion notation may be causing!
$endgroup$
– Namaste
Apr 18 '13 at 1:39
$begingroup$
don't worry, I understand it perfectly, thank you very much for your answer.
$endgroup$
– Diane Vanderwaif
Apr 18 '13 at 2:09
add a comment |
$begingroup$
Be careful: $$quad mathbb Z_{10} times mathbb Z_2 notcong mathbb Z_{20}; quadtext{since};gcd(2, 10) = 2neq 1$$
Use the facts that
- $U(44)= U(4cdot 11)cong U(4)times U(11)$, since $4$ and $11$ are relatively prime.
- $U(4)cong mathbb Z_2$ and $U(11)cong mathbb Z_{10}$.
To conclude that $$U(44) cong mathbb Z_{2}times mathbb Z_{10}$$
answered Apr 18 '13 at 0:13
NamasteNamaste
1
$endgroup$
Be careful: $$quad mathbb Z_{10} times mathbb Z_2 notcong mathbb Z_{20}; quadtext{since};gcd(2, 10) = 2neq 1$$
Use the facts that
- $U(44)= U(4cdot 11)cong U(4)times U(11)$, since $4$ and $11$ are relatively prime.
- $U(4)cong mathbb Z_2$ and $U(11)cong mathbb Z_{10}$.
To conclude that $$U(44) cong mathbb Z_{2}times mathbb Z_{10}$$
answered Apr 18 '13 at 0:13
NamasteNamaste
1
edited Apr 18 '13 at 0:29
answered Apr 18 '13 at 0:13
NamasteNamaste
1
answered Apr 18 '13 at 0:13
NamasteNamaste
1
answered Apr 18 '13 at 0:13
NamasteNamaste
1
1
$begingroup$
In my answer, $times$ denotes what you might know as the external direct product: $otimes$ or in the case of $mathbb Z_n times mathbb Z_m$ it denotes what some call the external direct sum: i.e., $mathbb Z_{10} otimes mathbb Z_2 notcong mathbb Z_{20},$ etc...with the conclusion that $U_{44} cong mathbb Z_2 otimes mathbb Z_{10}$. I apologize for any confusion notation may be causing!
$endgroup$
– Namaste
Apr 18 '13 at 1:39
$begingroup$
don't worry, I understand it perfectly, thank you very much for your answer.
$endgroup$
– Diane Vanderwaif
Apr 18 '13 at 2:09
add a comment |
$begingroup$
In my answer, $times$ denotes what you might know as the external direct product: $otimes$ or in the case of $mathbb Z_n times mathbb Z_m$ it denotes what some call the external direct sum: i.e., $mathbb Z_{10} otimes mathbb Z_2 notcong mathbb Z_{20},$ etc...with the conclusion that $U_{44} cong mathbb Z_2 otimes mathbb Z_{10}$. I apologize for any confusion notation may be causing!
$endgroup$
– Namaste
Apr 18 '13 at 1:39
$begingroup$
don't worry, I understand it perfectly, thank you very much for your answer.
$endgroup$
– Diane Vanderwaif
Apr 18 '13 at 2:09
$begingroup$
In my answer, $times$ denotes what you might know as the external direct product: $otimes$ or in the case of $mathbb Z_n times mathbb Z_m$ it denotes what some call the external direct sum: i.e., $mathbb Z_{10} otimes mathbb Z_2 notcong mathbb Z_{20},$ etc...with the conclusion that $U_{44} cong mathbb Z_2 otimes mathbb Z_{10}$. I apologize for any confusion notation may be causing!
$endgroup$
– Namaste
Apr 18 '13 at 1:39
$begingroup$
In my answer, $times$ denotes what you might know as the external direct product: $otimes$ or in the case of $mathbb Z_n times mathbb Z_m$ it denotes what some call the external direct sum: i.e., $mathbb Z_{10} otimes mathbb Z_2 notcong mathbb Z_{20},$ etc...with the conclusion that $U_{44} cong mathbb Z_2 otimes mathbb Z_{10}$. I apologize for any confusion notation may be causing!
$endgroup$
– Namaste
Apr 18 '13 at 1:39
$begingroup$
don't worry, I understand it perfectly, thank you very much for your answer.
$endgroup$
– Diane Vanderwaif
Apr 18 '13 at 2:09
$begingroup$
don't worry, I understand it perfectly, thank you very much for your answer.
$endgroup$
– Diane Vanderwaif
Apr 18 '13 at 2:09
add a comment |
$begingroup$
Hint: List the elements and compute their order. From this it should be quite clear.
answered Apr 18 '13 at 0:03
IslandsIslands
457213
$endgroup$
add a comment |
$begingroup$
Hint: List the elements and compute their order. From this it should be quite clear.
answered Apr 18 '13 at 0:03
IslandsIslands
457213
$endgroup$
add a comment |
$begingroup$
Hint: List the elements and compute their order. From this it should be quite clear.
answered Apr 18 '13 at 0:03
IslandsIslands
457213
$endgroup$
Hint: List the elements and compute their order. From this it should be quite clear.
answered Apr 18 '13 at 0:03
IslandsIslands
457213
answered Apr 18 '13 at 0:03
IslandsIslands
457213
answered Apr 18 '13 at 0:03
IslandsIslands
457213
answered Apr 18 '13 at 0:03
IslandsIslands
457213
457213
add a comment |
add a comment |
$begingroup$
In general, for $n=p_1^{e_1}cdots p_k^{e_k}$, $U_ncong(mathbb{Z}/p_1^{e_1}mathbb{Z})^timestimescdotstimes(mathbb{Z}/p_k^{e_k}mathbb{Z})^times$ by the Chinese Remainder Theorem.
answered Apr 18 '13 at 0:07
Warren MooreWarren Moore
3,2731128
$endgroup$
add a comment |
$begingroup$
In general, for $n=p_1^{e_1}cdots p_k^{e_k}$, $U_ncong(mathbb{Z}/p_1^{e_1}mathbb{Z})^timestimescdotstimes(mathbb{Z}/p_k^{e_k}mathbb{Z})^times$ by the Chinese Remainder Theorem.
answered Apr 18 '13 at 0:07
Warren MooreWarren Moore
3,2731128
$endgroup$
add a comment |
$begingroup$
In general, for $n=p_1^{e_1}cdots p_k^{e_k}$, $U_ncong(mathbb{Z}/p_1^{e_1}mathbb{Z})^timestimescdotstimes(mathbb{Z}/p_k^{e_k}mathbb{Z})^times$ by the Chinese Remainder Theorem.
answered Apr 18 '13 at 0:07
Warren MooreWarren Moore
3,2731128
$endgroup$
In general, for $n=p_1^{e_1}cdots p_k^{e_k}$, $U_ncong(mathbb{Z}/p_1^{e_1}mathbb{Z})^timestimescdotstimes(mathbb{Z}/p_k^{e_k}mathbb{Z})^times$ by the Chinese Remainder Theorem.
answered Apr 18 '13 at 0:07
Warren MooreWarren Moore
3,2731128
answered Apr 18 '13 at 0:07
Warren MooreWarren Moore
3,2731128
answered Apr 18 '13 at 0:07
Warren MooreWarren Moore
3,2731128
answered Apr 18 '13 at 0:07
Warren MooreWarren Moore
3,2731128
3,2731128
add a comment |
add a comment |
$begingroup$
Note that $U(44)= U(11*4)cong {mathbb Z_{10}} oplus {mathbb Z_2}$, since $gcd(11,4)=1.$
In general, if $gcd(a,b)=1$, then $U(ab)cong {mathbb Z_a}oplus{mathbb Z_b}$
answered Apr 18 '13 at 0:04
lukeluke
969819
$endgroup$
add a comment |
$begingroup$
Note that $U(44)= U(11*4)cong {mathbb Z_{10}} oplus {mathbb Z_2}$, since $gcd(11,4)=1.$
In general, if $gcd(a,b)=1$, then $U(ab)cong {mathbb Z_a}oplus{mathbb Z_b}$
answered Apr 18 '13 at 0:04
lukeluke
969819
$endgroup$
add a comment |
$begingroup$
Note that $U(44)= U(11*4)cong {mathbb Z_{10}} oplus {mathbb Z_2}$, since $gcd(11,4)=1.$
In general, if $gcd(a,b)=1$, then $U(ab)cong {mathbb Z_a}oplus{mathbb Z_b}$
answered Apr 18 '13 at 0:04
lukeluke
969819
$endgroup$
Note that $U(44)= U(11*4)cong {mathbb Z_{10}} oplus {mathbb Z_2}$, since $gcd(11,4)=1.$
In general, if $gcd(a,b)=1$, then $U(ab)cong {mathbb Z_a}oplus{mathbb Z_b}$
answered Apr 18 '13 at 0:04
lukeluke
969819
edited Apr 18 '13 at 0:26
answered Apr 18 '13 at 0:04
lukeluke
969819
answered Apr 18 '13 at 0:04
lukeluke
969819
answered Apr 18 '13 at 0:04
lukeluke
969819
969819
add a comment |
add a comment |
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$begingroup$
Actually, begin{align*} Bbb{Z}/20Bbb{Z}&congBbb{Z}/4Bbb{Z}timesBbb{Z}/5Bbb{Z}\ ¬congBbb{Z}/10Bbb{Z}timesBbb{Z}/2Bbb{Z}. end{align*} ($gcd(10,2) = 2neq 1$)
$endgroup$
– Stahl
Apr 18 '13 at 0:22