Mixing
Determining all homomorphisms from $Bbb Z_n times Bbb Z_m$ to itself
$begingroup$
I know how to determine all homomorphisms from $defZ{Bbb Z}Z_n$ to $Z_m$ (and to itself, naturally), but I can't seem to find an approach towards determining all homomorphisms from $Z_n times Z_m$ to itself. For example, from $Z_2timesZ_2$ to itself.
group-theory group-homomorphism
edited Feb 2 at 3:47
Martin Sleziak
45k10122277
asked Jan 8 '17 at 7:38
theSongbirdtheSongbird
523418
$endgroup$
add a comment |
$begingroup$
I know how to determine all homomorphisms from $defZ{Bbb Z}Z_n$ to $Z_m$ (and to itself, naturally), but I can't seem to find an approach towards determining all homomorphisms from $Z_n times Z_m$ to itself. For example, from $Z_2timesZ_2$ to itself.
group-theory group-homomorphism
edited Feb 2 at 3:47
Martin Sleziak
45k10122277
asked Jan 8 '17 at 7:38
theSongbirdtheSongbird
523418
$endgroup$
add a comment |
$begingroup$
I know how to determine all homomorphisms from $defZ{Bbb Z}Z_n$ to $Z_m$ (and to itself, naturally), but I can't seem to find an approach towards determining all homomorphisms from $Z_n times Z_m$ to itself. For example, from $Z_2timesZ_2$ to itself.
group-theory group-homomorphism
edited Feb 2 at 3:47
Martin Sleziak
45k10122277
asked Jan 8 '17 at 7:38
theSongbirdtheSongbird
523418
$endgroup$
I know how to determine all homomorphisms from $defZ{Bbb Z}Z_n$ to $Z_m$ (and to itself, naturally), but I can't seem to find an approach towards determining all homomorphisms from $Z_n times Z_m$ to itself. For example, from $Z_2timesZ_2$ to itself.
group-theory group-homomorphism
group-theory group-homomorphism
edited Feb 2 at 3:47
Martin Sleziak
45k10122277
asked Jan 8 '17 at 7:38
theSongbirdtheSongbird
523418
edited Feb 2 at 3:47
Martin Sleziak
45k10122277
asked Jan 8 '17 at 7:38
theSongbirdtheSongbird
523418
edited Feb 2 at 3:47
Martin Sleziak
45k10122277
edited Feb 2 at 3:47
Martin Sleziak
45k10122277
edited Feb 2 at 3:47
Martin Sleziak
45k10122277
45k10122277
asked Jan 8 '17 at 7:38
theSongbirdtheSongbird
523418
asked Jan 8 '17 at 7:38
theSongbirdtheSongbird
523418
asked Jan 8 '17 at 7:38
theSongbirdtheSongbird
523418
523418
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
You can simply send the generator of $defZ{Bbb Z}Z/nZ$ to any element $x$ satisfying $nx=0$, and the generator of $defZ{Bbb Z}Z/mZ$ to any element $y$ satisfying $my=0$. Then of course $(a,b)$ maps to $ax+by$. The sets of allowed values for $x,y$ are easy to determine, but depend somewhat on possible common divisors of $n$ and $m$.
For instance for $(Z/2Z)times(Z/2Z)$, all elements satisfy $2x=0$, so you have $4^2=16$ different group endomorphisms.
answered Jan 8 '17 at 7:43
Marc van LeeuwenMarc van Leeuwen
88.8k5111230
$endgroup$
$begingroup$
Are you referring to Z/Zn as to a quotient group? If so, that's mauly not be what I'm searching for.
$endgroup$
– theSongbird
Jan 8 '17 at 8:22
$begingroup$
So if not a quotient group, what do you mean by $Z_n$, the $n$-adic integers? If that is the case you had probably better say it clearly in your question, because most people here will read $Z_n$ as "the integers modulo $n$" instead (as indeed I did).
$endgroup$
– Marc van Leeuwen
Jan 8 '17 at 9:05
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can simply send the generator of $defZ{Bbb Z}Z/nZ$ to any element $x$ satisfying $nx=0$, and the generator of $defZ{Bbb Z}Z/mZ$ to any element $y$ satisfying $my=0$. Then of course $(a,b)$ maps to $ax+by$. The sets of allowed values for $x,y$ are easy to determine, but depend somewhat on possible common divisors of $n$ and $m$.
For instance for $(Z/2Z)times(Z/2Z)$, all elements satisfy $2x=0$, so you have $4^2=16$ different group endomorphisms.
answered Jan 8 '17 at 7:43
Marc van LeeuwenMarc van Leeuwen
88.8k5111230
$endgroup$
$begingroup$
Are you referring to Z/Zn as to a quotient group? If so, that's mauly not be what I'm searching for.
$endgroup$
– theSongbird
Jan 8 '17 at 8:22
$begingroup$
So if not a quotient group, what do you mean by $Z_n$, the $n$-adic integers? If that is the case you had probably better say it clearly in your question, because most people here will read $Z_n$ as "the integers modulo $n$" instead (as indeed I did).
$endgroup$
– Marc van Leeuwen
Jan 8 '17 at 9:05
add a comment |
$begingroup$
You can simply send the generator of $defZ{Bbb Z}Z/nZ$ to any element $x$ satisfying $nx=0$, and the generator of $defZ{Bbb Z}Z/mZ$ to any element $y$ satisfying $my=0$. Then of course $(a,b)$ maps to $ax+by$. The sets of allowed values for $x,y$ are easy to determine, but depend somewhat on possible common divisors of $n$ and $m$.
For instance for $(Z/2Z)times(Z/2Z)$, all elements satisfy $2x=0$, so you have $4^2=16$ different group endomorphisms.
answered Jan 8 '17 at 7:43
Marc van LeeuwenMarc van Leeuwen
88.8k5111230
$endgroup$
$begingroup$
Are you referring to Z/Zn as to a quotient group? If so, that's mauly not be what I'm searching for.
$endgroup$
– theSongbird
Jan 8 '17 at 8:22
$begingroup$
So if not a quotient group, what do you mean by $Z_n$, the $n$-adic integers? If that is the case you had probably better say it clearly in your question, because most people here will read $Z_n$ as "the integers modulo $n$" instead (as indeed I did).
$endgroup$
– Marc van Leeuwen
Jan 8 '17 at 9:05
add a comment |
$begingroup$
You can simply send the generator of $defZ{Bbb Z}Z/nZ$ to any element $x$ satisfying $nx=0$, and the generator of $defZ{Bbb Z}Z/mZ$ to any element $y$ satisfying $my=0$. Then of course $(a,b)$ maps to $ax+by$. The sets of allowed values for $x,y$ are easy to determine, but depend somewhat on possible common divisors of $n$ and $m$.
For instance for $(Z/2Z)times(Z/2Z)$, all elements satisfy $2x=0$, so you have $4^2=16$ different group endomorphisms.
answered Jan 8 '17 at 7:43
Marc van LeeuwenMarc van Leeuwen
88.8k5111230
$endgroup$
You can simply send the generator of $defZ{Bbb Z}Z/nZ$ to any element $x$ satisfying $nx=0$, and the generator of $defZ{Bbb Z}Z/mZ$ to any element $y$ satisfying $my=0$. Then of course $(a,b)$ maps to $ax+by$. The sets of allowed values for $x,y$ are easy to determine, but depend somewhat on possible common divisors of $n$ and $m$.
For instance for $(Z/2Z)times(Z/2Z)$, all elements satisfy $2x=0$, so you have $4^2=16$ different group endomorphisms.
answered Jan 8 '17 at 7:43
Marc van LeeuwenMarc van Leeuwen
88.8k5111230
edited Jan 8 '17 at 7:48
answered Jan 8 '17 at 7:43
Marc van LeeuwenMarc van Leeuwen
88.8k5111230
answered Jan 8 '17 at 7:43
Marc van LeeuwenMarc van Leeuwen
88.8k5111230
answered Jan 8 '17 at 7:43
Marc van LeeuwenMarc van Leeuwen
88.8k5111230
88.8k5111230
$begingroup$
Are you referring to Z/Zn as to a quotient group? If so, that's mauly not be what I'm searching for.
$endgroup$
– theSongbird
Jan 8 '17 at 8:22
$begingroup$
So if not a quotient group, what do you mean by $Z_n$, the $n$-adic integers? If that is the case you had probably better say it clearly in your question, because most people here will read $Z_n$ as "the integers modulo $n$" instead (as indeed I did).
$endgroup$
– Marc van Leeuwen
Jan 8 '17 at 9:05
add a comment |
$begingroup$
Are you referring to Z/Zn as to a quotient group? If so, that's mauly not be what I'm searching for.
$endgroup$
– theSongbird
Jan 8 '17 at 8:22
$begingroup$
So if not a quotient group, what do you mean by $Z_n$, the $n$-adic integers? If that is the case you had probably better say it clearly in your question, because most people here will read $Z_n$ as "the integers modulo $n$" instead (as indeed I did).
$endgroup$
– Marc van Leeuwen
Jan 8 '17 at 9:05
$begingroup$
Are you referring to Z/Zn as to a quotient group? If so, that's mauly not be what I'm searching for.
$endgroup$
– theSongbird
Jan 8 '17 at 8:22
$begingroup$
Are you referring to Z/Zn as to a quotient group? If so, that's mauly not be what I'm searching for.
$endgroup$
– theSongbird
Jan 8 '17 at 8:22
$begingroup$
So if not a quotient group, what do you mean by $Z_n$, the $n$-adic integers? If that is the case you had probably better say it clearly in your question, because most people here will read $Z_n$ as "the integers modulo $n$" instead (as indeed I did).
$endgroup$
– Marc van Leeuwen
Jan 8 '17 at 9:05
$begingroup$
So if not a quotient group, what do you mean by $Z_n$, the $n$-adic integers? If that is the case you had probably better say it clearly in your question, because most people here will read $Z_n$ as "the integers modulo $n$" instead (as indeed I did).
$endgroup$
– Marc van Leeuwen
Jan 8 '17 at 9:05
add a comment |
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