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Let G be $\underbrace{\mathbb Z_p\times\dots\times \mathbb Z_p}_{n \text{ times}}$. 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 $\mathbb Z_p\times\mathbb Z_p$ but I don't know if it's enough for $\mathbb Z_p\times \mathbb Z_p\times\dots\times\mathbb Z_p$ (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.

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    $\begingroup$ Use $\underbrace{X\times\dots\times X}_{n \text{ times}}$ for $\underbrace{X\times\dots\times X}_{n \text{ times}}$. $\endgroup$ – Shaun Jul 29 '18 at 12:03
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    $\begingroup$ What does the notation $A(G)$ mean here? $\endgroup$ – Henning Makholm Jul 29 '18 at 12:17
  • $\begingroup$ $A(G)$ is the automorphism group of G $\endgroup$ – user579852 Jul 29 '18 at 12:19
  • $\begingroup$ You say a vector space of $\mathbb{Z}_p$, but you tag pro-p-groups. By that notation do you mean the $p$-adic integers or the integers modulo $p$? $\endgroup$ – Henrique Augusto Souza Jul 29 '18 at 15:45
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    $\begingroup$ @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. $\endgroup$ – Andreas Blass Jul 29 '18 at 16:12
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Supposing that $\mathbb{Z}_p$ means the $p$-adic integers and you are asking about continuous (additive) group automorphisms, let's first note that any continuous group automorphism of $(\mathbb{Z}_p)^n$ is also a continuous $\mathbb{Z}_p$-module automorphism (as we can see the $\mathbb{Z}_p$ action as limits of $\mathbb{Z}$ actions). So we have:

$$A((\mathbb{Z}_p)^n) = A_{\mathbb{Z}_p}((\mathbb{Z}_p)^n)$$

But now, $(\mathbb{Z}_p)^n$ is a free $\mathbb{Z}_p$-module of rank $n$, in the sense that it behaves exactly like a vector space over a field (even though $\mathbb{Z}_p$ is not a field). Moreover, fixing a $\mathbb{Z}_p$-basis of $(\mathbb{Z}_p)^n$ (for example, the canonical one), every $\mathbb{Z}_p$-homomorphism of $(\mathbb{Z}_p)^n$ 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 $\mathbb{Z}_p$), and the matrix representation preserves automorphism compositions. This is enough to prove that

$$A_{\mathbb{Z}_p}((\mathbb{Z}_p)^n) \simeq \text{GL}_n(\mathbb{Z}_p)$$

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 $\mathbb{Z}_p$ means the integers modulo $p$, notice that $(\mathbb{Z}_p)^n$ is a vector space over $\mathbb{Z}_p$ of dimension $n$, and that the group automorphisms are also linear transformations (seeing the $\mathbb{Z}_p$ action as a quotient of a $\mathbb{Z}$ action). So $A((\mathbb{Z}_p)^n)$ is equal to the space of all invertible linear transformations os $(\mathbb{Z}_p)^n$ to itself. The rest of the proof is similar to the pro-p case.

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  • $\begingroup$ thank you for your answer I am sorry the meaning is modulo p $\endgroup$ – user579852 Jul 30 '18 at 14:48
  • $\begingroup$ so how does the prove for modulo p looks? $\endgroup$ – user579852 Jul 31 '18 at 12:01

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