Conditions for property of function $Bbb Z/mBbb Z to Bbb Z/nBbb Z$
$begingroup$
I've added a red star next to the line I don't understand, I also don't understand where the condition came from.
abstract-algebra number-theory
edited Dec 19 '18 at 14:23
Christoph
12.4k1642
asked Dec 19 '18 at 13:49
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
add a comment |
$begingroup$
I've added a red star next to the line I don't understand, I also don't understand where the condition came from.
abstract-algebra number-theory
edited Dec 19 '18 at 14:23
Christoph
12.4k1642
asked Dec 19 '18 at 13:49
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
1
$begingroup$
There are three steps in the line you highlighted, which one is not clear to you?
$endgroup$
– Christoph
Dec 19 '18 at 14:31
$begingroup$
The second two expressions
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 17:16
$begingroup$
@4M4D3U5 M0Z4RT, which book this question?
$endgroup$
– Mustafa
Dec 20 '18 at 9:23
add a comment |
$begingroup$
I've added a red star next to the line I don't understand, I also don't understand where the condition came from.
abstract-algebra number-theory
edited Dec 19 '18 at 14:23
Christoph
12.4k1642
asked Dec 19 '18 at 13:49
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
I've added a red star next to the line I don't understand, I also don't understand where the condition came from.
abstract-algebra number-theory
abstract-algebra number-theory
edited Dec 19 '18 at 14:23
Christoph
12.4k1642
asked Dec 19 '18 at 13:49
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
edited Dec 19 '18 at 14:23
Christoph
12.4k1642
asked Dec 19 '18 at 13:49
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
edited Dec 19 '18 at 14:23
Christoph
12.4k1642
edited Dec 19 '18 at 14:23
Christoph
12.4k1642
edited Dec 19 '18 at 14:23
Christoph
12.4k1642
12.4k1642
asked Dec 19 '18 at 13:49
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
asked Dec 19 '18 at 13:49
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
asked Dec 19 '18 at 13:49
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
386
1
$begingroup$
There are three steps in the line you highlighted, which one is not clear to you?
$endgroup$
– Christoph
Dec 19 '18 at 14:31
$begingroup$
The second two expressions
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 17:16
$begingroup$
@4M4D3U5 M0Z4RT, which book this question?
$endgroup$
– Mustafa
Dec 20 '18 at 9:23
add a comment |
1
$begingroup$
There are three steps in the line you highlighted, which one is not clear to you?
$endgroup$
– Christoph
Dec 19 '18 at 14:31
$begingroup$
The second two expressions
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 17:16
$begingroup$
@4M4D3U5 M0Z4RT, which book this question?
$endgroup$
– Mustafa
Dec 20 '18 at 9:23
1
1
$begingroup$
There are three steps in the line you highlighted, which one is not clear to you?
$endgroup$
– Christoph
Dec 19 '18 at 14:31
$begingroup$
There are three steps in the line you highlighted, which one is not clear to you?
$endgroup$
– Christoph
Dec 19 '18 at 14:31
$begingroup$
The second two expressions
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 17:16
$begingroup$
The second two expressions
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 17:16
$begingroup$
@4M4D3U5 M0Z4RT, which book this question?
$endgroup$
– Mustafa
Dec 20 '18 at 9:23
$begingroup$
@4M4D3U5 M0Z4RT, which book this question?
$endgroup$
– Mustafa
Dec 20 '18 at 9:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The remainder when you divide $0$ by $n$ is $0$ so $varphi(0) = 0$. You know that $m equiv 0 mod m$ so $varphi(m)=0$. Now you just write $m$ as $1+_mldots+_m 1$ ($m$ times) so you have $varphi(1+_mldots+_m 1) = 0$. Assuming the homomorphism property you get $varphi(1) +_n ldots +_n varphi(1)=0$. Since dividing 1 by $n$ you always get a remainder of $1$ as long as $n>1$, $varphi(1) = 1$. That means that $1 +_n ldots +_n 1 = 0$. This sum is in the group $mathbb{Z}/nmathbb{Z}$ but we still have $m$ summands so this says that the $m equiv 0 mod n$ which means $m$ must be a multiple of $n$.
answered Dec 19 '18 at 14:18
XiaopaiXiaopai
112
$endgroup$
add a comment |
$begingroup$
Remember that in $mathbb{Z}/nmathbb{Z}$, $underbrace{1+_ncdots +_n1}_{n}=0$. If you want $varphi(k+_m l)=varphi(k)+_nvarphi(l)$ (because you want a group homomorphism), in particular you need it for $k=l=1$. The reminder of devision of $1$ by any number greater than $1$ is again $1$, so that is where the equation
$$0=varphi(0)=varphi(underbrace{1+_ncdots +_n1}_{n})=underbrace{1+_mcdots +_m1}_{n}$$
comes frome. Adding $1$'s in $mathbb{Z}/nmathbb{Z}$ only sum up to $0$ if the times you add $1$ is a multiple of $n$. In this case, $m$ must be a multiple of $n$, which is the same as saying that $n$ must divide $m$.
Next, I'll show that this condition is sufficient. You can think of elements of $mathbb{Z}/mmathbb{Z}$ as equivalence classes of integers with the same reminder on the division by $m$, so take $overline{k}=k+rm$ and $overline{l}=l+sm$. Since $m=hn$, we have $overline{k}=k+rhn$ and $overline{l}=l+shn$. The homomorphism $varphi$ is just sending each element of $mathbb{Z}/mmathbb{Z}$ to itself modulo $n$. Hence, denoting equivalence classes by $_m$,
$varphi([overline{k}]_m+_m[overline{l}]_m)=varphi([overline{k}+overline{l}]_m)=[overline{k}+overline{l}]_n=[k+rhn+l+shn]_n=[k+l]_n=k+_n l$
I've used that the sum is not dependent on the representative of the equivalence, and the clear fact that taking the reminder of the division by $n$ doesn't depend on the representative.
answered Dec 19 '18 at 14:12
JaviJavi
2,7202827
$endgroup$
$begingroup$
How would i show that it is sufficient then?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 15:23
$begingroup$
@4M4D3U5M0Z4RT I've added it to the answer
$endgroup$
– Javi
Dec 19 '18 at 17:00
add a comment |
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2 Answers
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$begingroup$
The remainder when you divide $0$ by $n$ is $0$ so $varphi(0) = 0$. You know that $m equiv 0 mod m$ so $varphi(m)=0$. Now you just write $m$ as $1+_mldots+_m 1$ ($m$ times) so you have $varphi(1+_mldots+_m 1) = 0$. Assuming the homomorphism property you get $varphi(1) +_n ldots +_n varphi(1)=0$. Since dividing 1 by $n$ you always get a remainder of $1$ as long as $n>1$, $varphi(1) = 1$. That means that $1 +_n ldots +_n 1 = 0$. This sum is in the group $mathbb{Z}/nmathbb{Z}$ but we still have $m$ summands so this says that the $m equiv 0 mod n$ which means $m$ must be a multiple of $n$.
answered Dec 19 '18 at 14:18
XiaopaiXiaopai
112
$endgroup$
add a comment |
$begingroup$
The remainder when you divide $0$ by $n$ is $0$ so $varphi(0) = 0$. You know that $m equiv 0 mod m$ so $varphi(m)=0$. Now you just write $m$ as $1+_mldots+_m 1$ ($m$ times) so you have $varphi(1+_mldots+_m 1) = 0$. Assuming the homomorphism property you get $varphi(1) +_n ldots +_n varphi(1)=0$. Since dividing 1 by $n$ you always get a remainder of $1$ as long as $n>1$, $varphi(1) = 1$. That means that $1 +_n ldots +_n 1 = 0$. This sum is in the group $mathbb{Z}/nmathbb{Z}$ but we still have $m$ summands so this says that the $m equiv 0 mod n$ which means $m$ must be a multiple of $n$.
answered Dec 19 '18 at 14:18
XiaopaiXiaopai
112
$endgroup$
add a comment |
$begingroup$
The remainder when you divide $0$ by $n$ is $0$ so $varphi(0) = 0$. You know that $m equiv 0 mod m$ so $varphi(m)=0$. Now you just write $m$ as $1+_mldots+_m 1$ ($m$ times) so you have $varphi(1+_mldots+_m 1) = 0$. Assuming the homomorphism property you get $varphi(1) +_n ldots +_n varphi(1)=0$. Since dividing 1 by $n$ you always get a remainder of $1$ as long as $n>1$, $varphi(1) = 1$. That means that $1 +_n ldots +_n 1 = 0$. This sum is in the group $mathbb{Z}/nmathbb{Z}$ but we still have $m$ summands so this says that the $m equiv 0 mod n$ which means $m$ must be a multiple of $n$.
answered Dec 19 '18 at 14:18
XiaopaiXiaopai
112
$endgroup$
The remainder when you divide $0$ by $n$ is $0$ so $varphi(0) = 0$. You know that $m equiv 0 mod m$ so $varphi(m)=0$. Now you just write $m$ as $1+_mldots+_m 1$ ($m$ times) so you have $varphi(1+_mldots+_m 1) = 0$. Assuming the homomorphism property you get $varphi(1) +_n ldots +_n varphi(1)=0$. Since dividing 1 by $n$ you always get a remainder of $1$ as long as $n>1$, $varphi(1) = 1$. That means that $1 +_n ldots +_n 1 = 0$. This sum is in the group $mathbb{Z}/nmathbb{Z}$ but we still have $m$ summands so this says that the $m equiv 0 mod n$ which means $m$ must be a multiple of $n$.
answered Dec 19 '18 at 14:18
XiaopaiXiaopai
112
answered Dec 19 '18 at 14:18
XiaopaiXiaopai
112
answered Dec 19 '18 at 14:18
XiaopaiXiaopai
112
answered Dec 19 '18 at 14:18
XiaopaiXiaopai
112
112
add a comment |
add a comment |
$begingroup$
Remember that in $mathbb{Z}/nmathbb{Z}$, $underbrace{1+_ncdots +_n1}_{n}=0$. If you want $varphi(k+_m l)=varphi(k)+_nvarphi(l)$ (because you want a group homomorphism), in particular you need it for $k=l=1$. The reminder of devision of $1$ by any number greater than $1$ is again $1$, so that is where the equation
$$0=varphi(0)=varphi(underbrace{1+_ncdots +_n1}_{n})=underbrace{1+_mcdots +_m1}_{n}$$
comes frome. Adding $1$'s in $mathbb{Z}/nmathbb{Z}$ only sum up to $0$ if the times you add $1$ is a multiple of $n$. In this case, $m$ must be a multiple of $n$, which is the same as saying that $n$ must divide $m$.
Next, I'll show that this condition is sufficient. You can think of elements of $mathbb{Z}/mmathbb{Z}$ as equivalence classes of integers with the same reminder on the division by $m$, so take $overline{k}=k+rm$ and $overline{l}=l+sm$. Since $m=hn$, we have $overline{k}=k+rhn$ and $overline{l}=l+shn$. The homomorphism $varphi$ is just sending each element of $mathbb{Z}/mmathbb{Z}$ to itself modulo $n$. Hence, denoting equivalence classes by $_m$,
$varphi([overline{k}]_m+_m[overline{l}]_m)=varphi([overline{k}+overline{l}]_m)=[overline{k}+overline{l}]_n=[k+rhn+l+shn]_n=[k+l]_n=k+_n l$
I've used that the sum is not dependent on the representative of the equivalence, and the clear fact that taking the reminder of the division by $n$ doesn't depend on the representative.
answered Dec 19 '18 at 14:12
JaviJavi
2,7202827
$endgroup$
$begingroup$
How would i show that it is sufficient then?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 15:23
$begingroup$
@4M4D3U5M0Z4RT I've added it to the answer
$endgroup$
– Javi
Dec 19 '18 at 17:00
add a comment |
$begingroup$
Remember that in $mathbb{Z}/nmathbb{Z}$, $underbrace{1+_ncdots +_n1}_{n}=0$. If you want $varphi(k+_m l)=varphi(k)+_nvarphi(l)$ (because you want a group homomorphism), in particular you need it for $k=l=1$. The reminder of devision of $1$ by any number greater than $1$ is again $1$, so that is where the equation
$$0=varphi(0)=varphi(underbrace{1+_ncdots +_n1}_{n})=underbrace{1+_mcdots +_m1}_{n}$$
comes frome. Adding $1$'s in $mathbb{Z}/nmathbb{Z}$ only sum up to $0$ if the times you add $1$ is a multiple of $n$. In this case, $m$ must be a multiple of $n$, which is the same as saying that $n$ must divide $m$.
Next, I'll show that this condition is sufficient. You can think of elements of $mathbb{Z}/mmathbb{Z}$ as equivalence classes of integers with the same reminder on the division by $m$, so take $overline{k}=k+rm$ and $overline{l}=l+sm$. Since $m=hn$, we have $overline{k}=k+rhn$ and $overline{l}=l+shn$. The homomorphism $varphi$ is just sending each element of $mathbb{Z}/mmathbb{Z}$ to itself modulo $n$. Hence, denoting equivalence classes by $_m$,
$varphi([overline{k}]_m+_m[overline{l}]_m)=varphi([overline{k}+overline{l}]_m)=[overline{k}+overline{l}]_n=[k+rhn+l+shn]_n=[k+l]_n=k+_n l$
I've used that the sum is not dependent on the representative of the equivalence, and the clear fact that taking the reminder of the division by $n$ doesn't depend on the representative.
answered Dec 19 '18 at 14:12
JaviJavi
2,7202827
$endgroup$
$begingroup$
How would i show that it is sufficient then?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 15:23
$begingroup$
@4M4D3U5M0Z4RT I've added it to the answer
$endgroup$
– Javi
Dec 19 '18 at 17:00
add a comment |
$begingroup$
Remember that in $mathbb{Z}/nmathbb{Z}$, $underbrace{1+_ncdots +_n1}_{n}=0$. If you want $varphi(k+_m l)=varphi(k)+_nvarphi(l)$ (because you want a group homomorphism), in particular you need it for $k=l=1$. The reminder of devision of $1$ by any number greater than $1$ is again $1$, so that is where the equation
$$0=varphi(0)=varphi(underbrace{1+_ncdots +_n1}_{n})=underbrace{1+_mcdots +_m1}_{n}$$
comes frome. Adding $1$'s in $mathbb{Z}/nmathbb{Z}$ only sum up to $0$ if the times you add $1$ is a multiple of $n$. In this case, $m$ must be a multiple of $n$, which is the same as saying that $n$ must divide $m$.
Next, I'll show that this condition is sufficient. You can think of elements of $mathbb{Z}/mmathbb{Z}$ as equivalence classes of integers with the same reminder on the division by $m$, so take $overline{k}=k+rm$ and $overline{l}=l+sm$. Since $m=hn$, we have $overline{k}=k+rhn$ and $overline{l}=l+shn$. The homomorphism $varphi$ is just sending each element of $mathbb{Z}/mmathbb{Z}$ to itself modulo $n$. Hence, denoting equivalence classes by $_m$,
$varphi([overline{k}]_m+_m[overline{l}]_m)=varphi([overline{k}+overline{l}]_m)=[overline{k}+overline{l}]_n=[k+rhn+l+shn]_n=[k+l]_n=k+_n l$
I've used that the sum is not dependent on the representative of the equivalence, and the clear fact that taking the reminder of the division by $n$ doesn't depend on the representative.
answered Dec 19 '18 at 14:12
JaviJavi
2,7202827
$endgroup$
Remember that in $mathbb{Z}/nmathbb{Z}$, $underbrace{1+_ncdots +_n1}_{n}=0$. If you want $varphi(k+_m l)=varphi(k)+_nvarphi(l)$ (because you want a group homomorphism), in particular you need it for $k=l=1$. The reminder of devision of $1$ by any number greater than $1$ is again $1$, so that is where the equation
$$0=varphi(0)=varphi(underbrace{1+_ncdots +_n1}_{n})=underbrace{1+_mcdots +_m1}_{n}$$
comes frome. Adding $1$'s in $mathbb{Z}/nmathbb{Z}$ only sum up to $0$ if the times you add $1$ is a multiple of $n$. In this case, $m$ must be a multiple of $n$, which is the same as saying that $n$ must divide $m$.
Next, I'll show that this condition is sufficient. You can think of elements of $mathbb{Z}/mmathbb{Z}$ as equivalence classes of integers with the same reminder on the division by $m$, so take $overline{k}=k+rm$ and $overline{l}=l+sm$. Since $m=hn$, we have $overline{k}=k+rhn$ and $overline{l}=l+shn$. The homomorphism $varphi$ is just sending each element of $mathbb{Z}/mmathbb{Z}$ to itself modulo $n$. Hence, denoting equivalence classes by $_m$,
$varphi([overline{k}]_m+_m[overline{l}]_m)=varphi([overline{k}+overline{l}]_m)=[overline{k}+overline{l}]_n=[k+rhn+l+shn]_n=[k+l]_n=k+_n l$
I've used that the sum is not dependent on the representative of the equivalence, and the clear fact that taking the reminder of the division by $n$ doesn't depend on the representative.
answered Dec 19 '18 at 14:12
JaviJavi
2,7202827
edited Dec 19 '18 at 17:06
answered Dec 19 '18 at 14:12
JaviJavi
2,7202827
answered Dec 19 '18 at 14:12
JaviJavi
2,7202827
answered Dec 19 '18 at 14:12
JaviJavi
2,7202827
2,7202827
$begingroup$
How would i show that it is sufficient then?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 15:23
$begingroup$
@4M4D3U5M0Z4RT I've added it to the answer
$endgroup$
– Javi
Dec 19 '18 at 17:00
add a comment |
$begingroup$
How would i show that it is sufficient then?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 15:23
$begingroup$
@4M4D3U5M0Z4RT I've added it to the answer
$endgroup$
– Javi
Dec 19 '18 at 17:00
$begingroup$
How would i show that it is sufficient then?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 15:23
$begingroup$
How would i show that it is sufficient then?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 15:23
$begingroup$
@4M4D3U5M0Z4RT I've added it to the answer
$endgroup$
– Javi
Dec 19 '18 at 17:00
$begingroup$
@4M4D3U5M0Z4RT I've added it to the answer
$endgroup$
– Javi
Dec 19 '18 at 17:00
add a comment |
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1
$begingroup$
There are three steps in the line you highlighted, which one is not clear to you?
$endgroup$
– Christoph
Dec 19 '18 at 14:31
$begingroup$
The second two expressions
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 19 '18 at 17:16
$begingroup$
@4M4D3U5 M0Z4RT, which book this question?
$endgroup$
– Mustafa
Dec 20 '18 at 9:23