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$\begingroup$

I am trying to find the composition series of the group $\mathbb F_{p^k}$, where $p$ is prime , and $k\ge 1$. From Jordan Hölder equation it has length $k$ , i am quite confused , It looks quite natural to write $${e}\subset \mathbb F_p \subset...........\mathbb F_{p^k}$$

And each of them are definitely the maximal subgroups ( my doubt is how do i follow normality of every subgroup and the abelian nature of factor groups ie $\mathbb F_{p^i}/\mathbb F_{p^{i-1}}$ . Thank you .

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    The entire group is abelian, hence so is any factor and all subgroups are normal.2012-12-04
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    @Tobias : Ok , but why is entire group abelian ? All i know is that group of prime order is cyclic.2012-12-04
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    Well, how have you had the group in question defined?2012-12-04
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    Dear @Theorem, Do you really mean $\mathbf{F}_{p^k}$, as in a finite field of cardinality $p^k$? The reason I ask because, if you're only interested in its structure as a group, it is just $(\mathbf{Z}/p\mathbf{Z})^k$. Is it possible that you mean $\mathbf{Z}/p^k\mathbf{Z}$ instead?2012-12-04
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    @Tobias : I think its the usual way of defining group structure . What do you suggest ?2012-12-04
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    The reason I ask is that it really ought to be obvious from the definition that this group is abelian. So I have no way to know what you mean by the "usual" way.2012-12-04
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    @KeenanKidwell : I am sorry that i don't have complete information, Is it valid if its field with $p^k$ elements ? Does it make sense .2012-12-04
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    @Tobias : its embarrassing , but even i don't know . I need help to understand what could it possibly be.2012-12-04
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    If you don't know what the group is, then there is no way for us to help you.2012-12-04
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    @Tobias : Its finite field with $p^k$ elements :)2012-12-04
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    So the additive group of the field is abelian by definition.2012-12-04

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