Mixing
Let G be $mathbb Z_ptimesdotstimes mathbb Z_p$ . Find A(G).
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
Let G be $underbracemathbb Z_ptimesdotstimes mathbb Z_p_n text times$. Find $A(G)$.
I know that $A(G)cong GL_n(mathbb Z_p)$.
I prove it by taking $varphi$ from $A(G)$ and show that it's like matrix multiplication, than I took $theta$ from $A(G)$ and I show that the composition
of those two is a again matrix multiplication. Than I show that because we take matrix from $A(G)$ then there kernl is trivial so the matrix need to be not reversible.
I don't know if this is good way or not, I know it's good for $mathbb Z_ptimesmathbb Z_p$ but I don't know if it's enough for $mathbb Z_ptimes mathbb Z_ptimesdotstimesmathbb Z_p$ (n times).
If you have other wat please help me, there is a hint to think about vector space $(mathbb Z_p)^n$ but I don't know it could help.
finite-groups p-groups pro-p-groups
edited Jul 29 at 12:16
Henning Makholm
225k16290516
asked Jul 29 at 11:45
user579852
305
add a comment |Â
up vote
3
down vote
favorite
Let G be $underbracemathbb Z_ptimesdotstimes mathbb Z_p_n text times$. Find $A(G)$.
I know that $A(G)cong GL_n(mathbb Z_p)$.
I prove it by taking $varphi$ from $A(G)$ and show that it's like matrix multiplication, than I took $theta$ from $A(G)$ and I show that the composition
of those two is a again matrix multiplication. Than I show that because we take matrix from $A(G)$ then there kernl is trivial so the matrix need to be not reversible.
I don't know if this is good way or not, I know it's good for $mathbb Z_ptimesmathbb Z_p$ but I don't know if it's enough for $mathbb Z_ptimes mathbb Z_ptimesdotstimesmathbb Z_p$ (n times).
If you have other wat please help me, there is a hint to think about vector space $(mathbb Z_p)^n$ but I don't know it could help.
finite-groups p-groups pro-p-groups
edited Jul 29 at 12:16
Henning Makholm
225k16290516
asked Jul 29 at 11:45
user579852
305
1
Use$underbraceXtimesdotstimes X_n text times$for $underbraceXtimesdotstimes X_n text times$.
– Shaun
Jul 29 at 12:03
4
What does the notation $A(G)$ mean here?
– Henning Makholm
Jul 29 at 12:17
$A(G)$ is the automorphism group of G
– user579852
Jul 29 at 12:19
You say a vector space of $mathbbZ_p$, but you tag pro-p-groups. By that notation do you mean the $p$-adic integers or the integers modulo $p$?
– Henrique Augusto Souza
Jul 29 at 15:45
2
@HenriqueAugustoSouza The tag "finite groups" indicates the other meaning of $mathbb Z_p$ and, in view of the level of the question, I'm inclined to think this is the OP's intention.
– Andreas Blass
Jul 29 at 16:12
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let G be $underbracemathbb Z_ptimesdotstimes mathbb Z_p_n text times$. Find $A(G)$.
I know that $A(G)cong GL_n(mathbb Z_p)$.
I prove it by taking $varphi$ from $A(G)$ and show that it's like matrix multiplication, than I took $theta$ from $A(G)$ and I show that the composition
of those two is a again matrix multiplication. Than I show that because we take matrix from $A(G)$ then there kernl is trivial so the matrix need to be not reversible.
I don't know if this is good way or not, I know it's good for $mathbb Z_ptimesmathbb Z_p$ but I don't know if it's enough for $mathbb Z_ptimes mathbb Z_ptimesdotstimesmathbb Z_p$ (n times).
If you have other wat please help me, there is a hint to think about vector space $(mathbb Z_p)^n$ but I don't know it could help.
finite-groups p-groups pro-p-groups
edited Jul 29 at 12:16
Henning Makholm
225k16290516
asked Jul 29 at 11:45
user579852
305
Let G be $underbracemathbb Z_ptimesdotstimes mathbb Z_p_n text times$. Find $A(G)$.
I know that $A(G)cong GL_n(mathbb Z_p)$.
I prove it by taking $varphi$ from $A(G)$ and show that it's like matrix multiplication, than I took $theta$ from $A(G)$ and I show that the composition
of those two is a again matrix multiplication. Than I show that because we take matrix from $A(G)$ then there kernl is trivial so the matrix need to be not reversible.
I don't know if this is good way or not, I know it's good for $mathbb Z_ptimesmathbb Z_p$ but I don't know if it's enough for $mathbb Z_ptimes mathbb Z_ptimesdotstimesmathbb Z_p$ (n times).
If you have other wat please help me, there is a hint to think about vector space $(mathbb Z_p)^n$ but I don't know it could help.
finite-groups p-groups pro-p-groups
edited Jul 29 at 12:16
Henning Makholm
225k16290516
asked Jul 29 at 11:45
user579852
305
edited Jul 29 at 12:16
Henning Makholm
225k16290516
edited Jul 29 at 12:16
Henning Makholm
225k16290516
edited Jul 29 at 12:16
Henning Makholm
225k16290516
225k16290516
asked Jul 29 at 11:45
user579852
305
asked Jul 29 at 11:45
user579852
305
asked Jul 29 at 11:45
user579852
305
305
1
Use$underbraceXtimesdotstimes X_n text times$for $underbraceXtimesdotstimes X_n text times$.
– Shaun
Jul 29 at 12:03
4
What does the notation $A(G)$ mean here?
– Henning Makholm
Jul 29 at 12:17
$A(G)$ is the automorphism group of G
– user579852
Jul 29 at 12:19
You say a vector space of $mathbbZ_p$, but you tag pro-p-groups. By that notation do you mean the $p$-adic integers or the integers modulo $p$?
– Henrique Augusto Souza
Jul 29 at 15:45
2
@HenriqueAugustoSouza The tag "finite groups" indicates the other meaning of $mathbb Z_p$ and, in view of the level of the question, I'm inclined to think this is the OP's intention.
– Andreas Blass
Jul 29 at 16:12
add a comment |Â
1
Use$underbraceXtimesdotstimes X_n text times$for $underbraceXtimesdotstimes X_n text times$.
– Shaun
Jul 29 at 12:03
4
What does the notation $A(G)$ mean here?
– Henning Makholm
Jul 29 at 12:17
$A(G)$ is the automorphism group of G
– user579852
Jul 29 at 12:19
You say a vector space of $mathbbZ_p$, but you tag pro-p-groups. By that notation do you mean the $p$-adic integers or the integers modulo $p$?
– Henrique Augusto Souza
Jul 29 at 15:45
2
@HenriqueAugustoSouza The tag "finite groups" indicates the other meaning of $mathbb Z_p$ and, in view of the level of the question, I'm inclined to think this is the OP's intention.
– Andreas Blass
Jul 29 at 16:12
1
1
Use
$underbraceXtimesdotstimes X_n text times$ for $underbraceXtimesdotstimes X_n text times$.– Shaun
Jul 29 at 12:03
Use
$underbraceXtimesdotstimes X_n text times$ for $underbraceXtimesdotstimes X_n text times$.– Shaun
Jul 29 at 12:03
4
4
What does the notation $A(G)$ mean here?
– Henning Makholm
Jul 29 at 12:17
What does the notation $A(G)$ mean here?
– Henning Makholm
Jul 29 at 12:17
$A(G)$ is the automorphism group of G
– user579852
Jul 29 at 12:19
$A(G)$ is the automorphism group of G
– user579852
Jul 29 at 12:19
You say a vector space of $mathbbZ_p$, but you tag pro-p-groups. By that notation do you mean the $p$-adic integers or the integers modulo $p$?
– Henrique Augusto Souza
Jul 29 at 15:45
You say a vector space of $mathbbZ_p$, but you tag pro-p-groups. By that notation do you mean the $p$-adic integers or the integers modulo $p$?
– Henrique Augusto Souza
Jul 29 at 15:45
2
2
@HenriqueAugustoSouza The tag "finite groups" indicates the other meaning of $mathbb Z_p$ and, in view of the level of the question, I'm inclined to think this is the OP's intention.
– Andreas Blass
Jul 29 at 16:12
@HenriqueAugustoSouza The tag "finite groups" indicates the other meaning of $mathbb Z_p$ and, in view of the level of the question, I'm inclined to think this is the OP's intention.
– Andreas Blass
Jul 29 at 16:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Supposing that $mathbbZ_p$ means the $p$-adic integers and you are asking about continuous (additive) group automorphisms, let's first note that any continuous group automorphism of $(mathbbZ_p)^n$ is also a continuous $mathbbZ_p$-module automorphism (as we can see the $mathbbZ_p$ action as limits of $mathbbZ$ actions). So we have:
$$A((mathbbZ_p)^n) = A_mathbbZ_p((mathbbZ_p)^n)$$
But now, $(mathbbZ_p)^n$ is a free $mathbbZ_p$-module of rank $n$, in the sense that it behaves exactly like a vector space over a field (even though $mathbbZ_p$ is not a field). Moreover, fixing a $mathbbZ_p$-basis of $(mathbbZ_p)^n$ (for example, the canonical one), every $mathbbZ_p$-homomorphism of $(mathbbZ_p)^n$ is uniquely determined by it's image on the basis and can be uniquely represented by a matrix. The homomorphism is an automorphism if and only if the matrix is invertible (it's determinant being a unit on $mathbbZ_p$), and the matrix representation preserves automorphism compositions. This is enough to prove that
$$A_mathbbZ_p((mathbbZ_p)^n) simeq textGL_n(mathbbZ_p)$$
The only tricky part here is showing that the matrix is invertible iff it's determinant is a unit (check Theorem 7.9 in Serge Lang's Algebra). But this isn't necessary to prove the isomorphism, it is just a remark.
If $mathbbZ_p$ means the integers modulo $p$, notice that $(mathbbZ_p)^n$ is a vector space over $mathbbZ_p$ of dimension $n$, and that the group automorphisms are also linear transformations (seeing the $mathbbZ_p$ action as a quotient of a $mathbbZ$ action). So $A((mathbbZ_p)^n)$ is equal to the space of all invertible linear transformations os $(mathbbZ_p)^n$ to itself. The rest of the proof is similar to the pro-p case.
answered Jul 29 at 15:59
Henrique Augusto Souza
1,226314
thank you for your answer I am sorry the meaning is modulo p
– user579852
Jul 30 at 14:48
so how does the prove for modulo p looks?
– user579852
Jul 31 at 12:01
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Supposing that $mathbbZ_p$ means the $p$-adic integers and you are asking about continuous (additive) group automorphisms, let's first note that any continuous group automorphism of $(mathbbZ_p)^n$ is also a continuous $mathbbZ_p$-module automorphism (as we can see the $mathbbZ_p$ action as limits of $mathbbZ$ actions). So we have:
$$A((mathbbZ_p)^n) = A_mathbbZ_p((mathbbZ_p)^n)$$
But now, $(mathbbZ_p)^n$ is a free $mathbbZ_p$-module of rank $n$, in the sense that it behaves exactly like a vector space over a field (even though $mathbbZ_p$ is not a field). Moreover, fixing a $mathbbZ_p$-basis of $(mathbbZ_p)^n$ (for example, the canonical one), every $mathbbZ_p$-homomorphism of $(mathbbZ_p)^n$ is uniquely determined by it's image on the basis and can be uniquely represented by a matrix. The homomorphism is an automorphism if and only if the matrix is invertible (it's determinant being a unit on $mathbbZ_p$), and the matrix representation preserves automorphism compositions. This is enough to prove that
$$A_mathbbZ_p((mathbbZ_p)^n) simeq textGL_n(mathbbZ_p)$$
The only tricky part here is showing that the matrix is invertible iff it's determinant is a unit (check Theorem 7.9 in Serge Lang's Algebra). But this isn't necessary to prove the isomorphism, it is just a remark.
If $mathbbZ_p$ means the integers modulo $p$, notice that $(mathbbZ_p)^n$ is a vector space over $mathbbZ_p$ of dimension $n$, and that the group automorphisms are also linear transformations (seeing the $mathbbZ_p$ action as a quotient of a $mathbbZ$ action). So $A((mathbbZ_p)^n)$ is equal to the space of all invertible linear transformations os $(mathbbZ_p)^n$ to itself. The rest of the proof is similar to the pro-p case.
answered Jul 29 at 15:59
Henrique Augusto Souza
1,226314
thank you for your answer I am sorry the meaning is modulo p
– user579852
Jul 30 at 14:48
so how does the prove for modulo p looks?
– user579852
Jul 31 at 12:01
add a comment |Â
up vote
1
down vote
Supposing that $mathbbZ_p$ means the $p$-adic integers and you are asking about continuous (additive) group automorphisms, let's first note that any continuous group automorphism of $(mathbbZ_p)^n$ is also a continuous $mathbbZ_p$-module automorphism (as we can see the $mathbbZ_p$ action as limits of $mathbbZ$ actions). So we have:
$$A((mathbbZ_p)^n) = A_mathbbZ_p((mathbbZ_p)^n)$$
But now, $(mathbbZ_p)^n$ is a free $mathbbZ_p$-module of rank $n$, in the sense that it behaves exactly like a vector space over a field (even though $mathbbZ_p$ is not a field). Moreover, fixing a $mathbbZ_p$-basis of $(mathbbZ_p)^n$ (for example, the canonical one), every $mathbbZ_p$-homomorphism of $(mathbbZ_p)^n$ is uniquely determined by it's image on the basis and can be uniquely represented by a matrix. The homomorphism is an automorphism if and only if the matrix is invertible (it's determinant being a unit on $mathbbZ_p$), and the matrix representation preserves automorphism compositions. This is enough to prove that
$$A_mathbbZ_p((mathbbZ_p)^n) simeq textGL_n(mathbbZ_p)$$
The only tricky part here is showing that the matrix is invertible iff it's determinant is a unit (check Theorem 7.9 in Serge Lang's Algebra). But this isn't necessary to prove the isomorphism, it is just a remark.
If $mathbbZ_p$ means the integers modulo $p$, notice that $(mathbbZ_p)^n$ is a vector space over $mathbbZ_p$ of dimension $n$, and that the group automorphisms are also linear transformations (seeing the $mathbbZ_p$ action as a quotient of a $mathbbZ$ action). So $A((mathbbZ_p)^n)$ is equal to the space of all invertible linear transformations os $(mathbbZ_p)^n$ to itself. The rest of the proof is similar to the pro-p case.
answered Jul 29 at 15:59
Henrique Augusto Souza
1,226314
thank you for your answer I am sorry the meaning is modulo p
– user579852
Jul 30 at 14:48
so how does the prove for modulo p looks?
– user579852
Jul 31 at 12:01
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Supposing that $mathbbZ_p$ means the $p$-adic integers and you are asking about continuous (additive) group automorphisms, let's first note that any continuous group automorphism of $(mathbbZ_p)^n$ is also a continuous $mathbbZ_p$-module automorphism (as we can see the $mathbbZ_p$ action as limits of $mathbbZ$ actions). So we have:
$$A((mathbbZ_p)^n) = A_mathbbZ_p((mathbbZ_p)^n)$$
But now, $(mathbbZ_p)^n$ is a free $mathbbZ_p$-module of rank $n$, in the sense that it behaves exactly like a vector space over a field (even though $mathbbZ_p$ is not a field). Moreover, fixing a $mathbbZ_p$-basis of $(mathbbZ_p)^n$ (for example, the canonical one), every $mathbbZ_p$-homomorphism of $(mathbbZ_p)^n$ is uniquely determined by it's image on the basis and can be uniquely represented by a matrix. The homomorphism is an automorphism if and only if the matrix is invertible (it's determinant being a unit on $mathbbZ_p$), and the matrix representation preserves automorphism compositions. This is enough to prove that
$$A_mathbbZ_p((mathbbZ_p)^n) simeq textGL_n(mathbbZ_p)$$
The only tricky part here is showing that the matrix is invertible iff it's determinant is a unit (check Theorem 7.9 in Serge Lang's Algebra). But this isn't necessary to prove the isomorphism, it is just a remark.
If $mathbbZ_p$ means the integers modulo $p$, notice that $(mathbbZ_p)^n$ is a vector space over $mathbbZ_p$ of dimension $n$, and that the group automorphisms are also linear transformations (seeing the $mathbbZ_p$ action as a quotient of a $mathbbZ$ action). So $A((mathbbZ_p)^n)$ is equal to the space of all invertible linear transformations os $(mathbbZ_p)^n$ to itself. The rest of the proof is similar to the pro-p case.
answered Jul 29 at 15:59
Henrique Augusto Souza
1,226314
Supposing that $mathbbZ_p$ means the $p$-adic integers and you are asking about continuous (additive) group automorphisms, let's first note that any continuous group automorphism of $(mathbbZ_p)^n$ is also a continuous $mathbbZ_p$-module automorphism (as we can see the $mathbbZ_p$ action as limits of $mathbbZ$ actions). So we have:
$$A((mathbbZ_p)^n) = A_mathbbZ_p((mathbbZ_p)^n)$$
But now, $(mathbbZ_p)^n$ is a free $mathbbZ_p$-module of rank $n$, in the sense that it behaves exactly like a vector space over a field (even though $mathbbZ_p$ is not a field). Moreover, fixing a $mathbbZ_p$-basis of $(mathbbZ_p)^n$ (for example, the canonical one), every $mathbbZ_p$-homomorphism of $(mathbbZ_p)^n$ is uniquely determined by it's image on the basis and can be uniquely represented by a matrix. The homomorphism is an automorphism if and only if the matrix is invertible (it's determinant being a unit on $mathbbZ_p$), and the matrix representation preserves automorphism compositions. This is enough to prove that
$$A_mathbbZ_p((mathbbZ_p)^n) simeq textGL_n(mathbbZ_p)$$
The only tricky part here is showing that the matrix is invertible iff it's determinant is a unit (check Theorem 7.9 in Serge Lang's Algebra). But this isn't necessary to prove the isomorphism, it is just a remark.
If $mathbbZ_p$ means the integers modulo $p$, notice that $(mathbbZ_p)^n$ is a vector space over $mathbbZ_p$ of dimension $n$, and that the group automorphisms are also linear transformations (seeing the $mathbbZ_p$ action as a quotient of a $mathbbZ$ action). So $A((mathbbZ_p)^n)$ is equal to the space of all invertible linear transformations os $(mathbbZ_p)^n$ to itself. The rest of the proof is similar to the pro-p case.
answered Jul 29 at 15:59
Henrique Augusto Souza
1,226314
edited Jul 29 at 16:05
answered Jul 29 at 15:59
Henrique Augusto Souza
1,226314
answered Jul 29 at 15:59
Henrique Augusto Souza
1,226314
answered Jul 29 at 15:59
Henrique Augusto Souza
1,226314
1,226314
thank you for your answer I am sorry the meaning is modulo p
– user579852
Jul 30 at 14:48
so how does the prove for modulo p looks?
– user579852
Jul 31 at 12:01
add a comment |Â
thank you for your answer I am sorry the meaning is modulo p
– user579852
Jul 30 at 14:48
so how does the prove for modulo p looks?
– user579852
Jul 31 at 12:01
thank you for your answer I am sorry the meaning is modulo p
– user579852
Jul 30 at 14:48
thank you for your answer I am sorry the meaning is modulo p
– user579852
Jul 30 at 14:48
so how does the prove for modulo p looks?
– user579852
Jul 31 at 12:01
so how does the prove for modulo p looks?
– user579852
Jul 31 at 12:01
add a comment |Â
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1
Use
$underbraceXtimesdotstimes X_n text times$for $underbraceXtimesdotstimes X_n text times$.– Shaun
Jul 29 at 12:03
4
What does the notation $A(G)$ mean here?
– Henning Makholm
Jul 29 at 12:17
$A(G)$ is the automorphism group of G
– user579852
Jul 29 at 12:19
You say a vector space of $mathbbZ_p$, but you tag pro-p-groups. By that notation do you mean the $p$-adic integers or the integers modulo $p$?
– Henrique Augusto Souza
Jul 29 at 15:45
2
@HenriqueAugustoSouza The tag "finite groups" indicates the other meaning of $mathbb Z_p$ and, in view of the level of the question, I'm inclined to think this is the OP's intention.
– Andreas Blass
Jul 29 at 16:12