Cardinality of $GL_n(Z_p^m)$

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Let $p$ be a prime number and $Z_p^m$ be the ring of integers modulo $p^m.$ Now one may consider a map $f:Z_p^m to Z_p$ which clearly induces a map $F:GL_n(Z_p^m) to GL_n(Z_p)$ because of the fact that for any integer $a,$ $gcd(a,p^m)=1$ iff $gcd(a,p)=1.$ Clearly $F$ is given by $F((a_ij)_n)=(f(a_ij))_n.$ It is also clear that $F$ is surjective. But I want to know the cardinality of $GL_n(Z_p^m),$ for this I have to know the cardinality of $Ker(F).$ I need some help to calculate the cardinality of $Ker(F).$ Note that class of the elements $1,p+1,2p+1,ldots,p^m-p+1$ in $Z_p$ is $1$ and class of the elements $0,p,2p,ldots,p^m-p$ in $Z_p$ is $0.$

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linear-algebra combinatorics matrices ring-theory finite-groups

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asked Jul 21 at 11:20

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Let $p$ be a prime number and $Z_p^m$ be the ring of integers modulo $p^m.$ Now one may consider a map $f:Z_p^m to Z_p$ which clearly induces a map $F:GL_n(Z_p^m) to GL_n(Z_p)$ because of the fact that for any integer $a,$ $gcd(a,p^m)=1$ iff $gcd(a,p)=1.$ Clearly $F$ is given by $F((a_ij)_n)=(f(a_ij))_n.$ It is also clear that $F$ is surjective. But I want to know the cardinality of $GL_n(Z_p^m),$ for this I have to know the cardinality of $Ker(F).$ I need some help to calculate the cardinality of $Ker(F).$ Note that class of the elements $1,p+1,2p+1,ldots,p^m-p+1$ in $Z_p$ is $1$ and class of the elements $0,p,2p,ldots,p^m-p$ in $Z_p$ is $0.$

Thanks

linear-algebra combinatorics matrices ring-theory finite-groups

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asked Jul 21 at 11:20

user371231

411410

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up vote
3
down vote

favorite

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up vote
3
down vote

favorite

2

2

Let $p$ be a prime number and $Z_p^m$ be the ring of integers modulo $p^m.$ Now one may consider a map $f:Z_p^m to Z_p$ which clearly induces a map $F:GL_n(Z_p^m) to GL_n(Z_p)$ because of the fact that for any integer $a,$ $gcd(a,p^m)=1$ iff $gcd(a,p)=1.$ Clearly $F$ is given by $F((a_ij)_n)=(f(a_ij))_n.$ It is also clear that $F$ is surjective. But I want to know the cardinality of $GL_n(Z_p^m),$ for this I have to know the cardinality of $Ker(F).$ I need some help to calculate the cardinality of $Ker(F).$ Note that class of the elements $1,p+1,2p+1,ldots,p^m-p+1$ in $Z_p$ is $1$ and class of the elements $0,p,2p,ldots,p^m-p$ in $Z_p$ is $0.$

Thanks

linear-algebra combinatorics matrices ring-theory finite-groups

share|cite|improve this question

asked Jul 21 at 11:20

user371231

411410

Let $p$ be a prime number and $Z_p^m$ be the ring of integers modulo $p^m.$ Now one may consider a map $f:Z_p^m to Z_p$ which clearly induces a map $F:GL_n(Z_p^m) to GL_n(Z_p)$ because of the fact that for any integer $a,$ $gcd(a,p^m)=1$ iff $gcd(a,p)=1.$ Clearly $F$ is given by $F((a_ij)_n)=(f(a_ij))_n.$ It is also clear that $F$ is surjective. But I want to know the cardinality of $GL_n(Z_p^m),$ for this I have to know the cardinality of $Ker(F).$ I need some help to calculate the cardinality of $Ker(F).$ Note that class of the elements $1,p+1,2p+1,ldots,p^m-p+1$ in $Z_p$ is $1$ and class of the elements $0,p,2p,ldots,p^m-p$ in $Z_p$ is $0.$

Thanks

linear-algebra combinatorics matrices ring-theory finite-groups

share|cite|improve this question

asked Jul 21 at 11:20

user371231

411410

share|cite|improve this question

share|cite|improve this question

asked Jul 21 at 11:20

user371231

411410

asked Jul 21 at 11:20

user371231

411410

asked Jul 21 at 11:20

user371231

411410

411410

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add a comment | 

1 Answer
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The kernel consists of those elements $MintextrmGL_n(Z_p^m)$ that are
congruent to the identity $I$ modulo $p$. There are $p^m-1$ choices
for each entry of $M$, the off-diagonal entries must lie in $pZ_p^m$
and the diagonal entries msu lie in $1+pZ_p^m$.

Therefore the kernel has order $p^(m-1)n^2$.

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answered Jul 21 at 11:27

Lord Shark the Unknown

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The kernel consists of those elements $MintextrmGL_n(Z_p^m)$ that are
congruent to the identity $I$ modulo $p$. There are $p^m-1$ choices
for each entry of $M$, the off-diagonal entries must lie in $pZ_p^m$
and the diagonal entries msu lie in $1+pZ_p^m$.

Therefore the kernel has order $p^(m-1)n^2$.

share|cite|improve this answer

answered Jul 21 at 11:27

Lord Shark the Unknown

85.2k950111

add a comment | 

up vote
2
down vote



accepted

The kernel consists of those elements $MintextrmGL_n(Z_p^m)$ that are
congruent to the identity $I$ modulo $p$. There are $p^m-1$ choices
for each entry of $M$, the off-diagonal entries must lie in $pZ_p^m$
and the diagonal entries msu lie in $1+pZ_p^m$.

Therefore the kernel has order $p^(m-1)n^2$.

share|cite|improve this answer

answered Jul 21 at 11:27

Lord Shark the Unknown

85.2k950111

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The kernel consists of those elements $MintextrmGL_n(Z_p^m)$ that are
congruent to the identity $I$ modulo $p$. There are $p^m-1$ choices
for each entry of $M$, the off-diagonal entries must lie in $pZ_p^m$
and the diagonal entries msu lie in $1+pZ_p^m$.

Therefore the kernel has order $p^(m-1)n^2$.

share|cite|improve this answer

answered Jul 21 at 11:27

Lord Shark the Unknown

85.2k950111

The kernel consists of those elements $MintextrmGL_n(Z_p^m)$ that are
congruent to the identity $I$ modulo $p$. There are $p^m-1$ choices
for each entry of $M$, the off-diagonal entries must lie in $pZ_p^m$
and the diagonal entries msu lie in $1+pZ_p^m$.

Therefore the kernel has order $p^(m-1)n^2$.

share|cite|improve this answer

answered Jul 21 at 11:27

Lord Shark the Unknown

85.2k950111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 11:27

Lord Shark the Unknown

85.2k950111

answered Jul 21 at 11:27

Lord Shark the Unknown

85.2k950111

answered Jul 21 at 11:27

Lord Shark the Unknown

85.2k950111

85.2k950111

add a comment | 

add a comment | 

 

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