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    When n is arbitrary but e_i =1 for 1 \leq i \leq n. We consider P to be of the form \mathbb{Z}_p\oplus...\mathbb{Z}_  p\oplus , so elements of this subgroup can be viewed as comprising a vector space of dimension n over the finite field \mathbb{F}_p of p elements . The automorphisms of this subgroup are therefore given by the invertible linear transformations, so

    Aut(P) \cong GL(n, \mathbb{F}_p) has order |Aut(P)| = (p^n -1)...(p^n - p^{n-1}) *

    I know that if you define d_k = max\{r|e_r = e_k\} and c_k = min\{r|e_r = e_k\}, then one has in particular d_k \geq k, c_k \leq k and,

    |Aut(P)| = \left( \displaystyle\prod_{k=1}^n p^{d_k} - p^{k-1} \right) \left( \displaystyle\prod_{j=1}^n (p^{e_j})^{n-d_j} \right) \left( \displaystyle\prod_{i=1}^n (p^{e_i -1})^{n - c_i +1} \right) , however I think this follows from showing that Aut(P) has order (p^n -1)...(p^n - p^{n-1}) .

    *Since my textbook just says "this is easily shown to have order...", I'm a little unsettled. How can I show it explicitly? I've had a few ideas, but nothing seems tangible, unless I'm missing something important, which I probably am.
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    What vectors can the first column of an invertible matrix over F_p be?

    Given the first column vector, what can the second column vector be?

    etc.
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    (Original post by RichE)
    What vectors can the first column of an invertible matrix over F_p be?

    Given the first column vector, what can the second column vector be?

    etc.
    Got it now, cheers.
 
 
 
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