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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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$\mathbb{F}_{p^k}$ is a finite field, which means it has an additive group and a multiplicative group. You can think of elements of $\mathbb{F}_{p^k}$ as polynomials $a_0+a_1x+\ldots +a_{k-1}x^{k-1}$, where the $a_i\in \mathbb{Z}_p$ and $x$ is an indeterminate element. (It's easy to see combinatorially that there are $p^k$ polynomials of this form.) The addition is done just like polynomial addition, that is, $$\left(a_0+a_1x+\ldots +a_{k-1}x^{k-1}\right)+\left(b_0+b_1x+\ldots +b_{k-1}\alpha^{k-1}\right)\\=(a_0+b_0)+(a_1+b_1)x+\ldots +(a_{k-1}+b_{k-1})x^{k-1}.$$ The multiplication is more complicated. When you make $\mathbb{F}_{p^k}$, you do it by picking some polynomial $m(x)$ of degree $k$ which is irreducible over the polynomial ring $\mathbb{Z}_p[x]$. Then you 'mod out' by $m(x)$ in $\mathbb{Z}_p[x]$. What this means is that when you multiply two elements $p(x)$ nd $q(x)$ of $\mathbb{F}_{p^k}$ written in the form given above, you take the remainder of $p(x)q(x)$ modulo $m(x)$. This is done simply by polynomial long division, which is exactly the same as normal long devision - you just keep dividing by $m(x)$ until you can't anymore, and whatever is left over is $p(x)q(x)$ in the desired polynomial form.

Anyway, that's a little introduction to finite fields to help you on your way. You can see that both the addition and multiplication are commutative operations. If you look up the definition of normality, I am sure this will ease your concerns. If you'd like some advice though, I think you should go back and relearn the basic definitions of all this stuff before you try to read about advanced things like Jordan-Holder (which, by the way, doesn't just apply to groups). It's going to be difficult to solve problems if you don't know what the objects you're working with are.

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    Ok , this is slightly confusing when you say that Jordan Hölder theorem is not for groups . http://mathworld.wolfram.com/Jordan-HoelderTheorem.html2012-12-05
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    @Theorem Jorden-Hölder is for groups, but it's also for other algebraic structures, namely modules. (I actually learned it for modules first.) This is probably important to whatever it is you're doing because you asked about the composition series for a *field*, not just for its additive group.2012-12-05