How to prove that every finite field of order $125$ has a subfield of order $25$. In general what is the strategy to attack such kind of problems?
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1See: http://math.stackexchange.com/questions/91087/subfields-of-finite-fields – JavaMan Nov 10 '12 at 05:04
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7You'll never prove it, it’s not true. – Lubin Nov 10 '12 at 05:32
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Consider objects of the form $x^5$, where $x$ ranges over the big field. – André Nicolas Nov 10 '12 at 05:34
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Consider a field $F$ with $q$ elements, $q$ being a power of some prime. Suppose $L$ is a field containing $F$, with $[L\colon F]=m$. Since $L$ is an $F$-vector space of dimension $m$, $|L|=q^m$. Thus $[\mathbb F_{125}\colon\mathbb F_5]=3$. Now do you see why there’s no field strictly between these two fields?
Lubin
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Thank you. I understand your argument. I didnt in the least feel this could be a false statement. – rTeja Nov 10 '12 at 07:39
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2You don't need to consider $\Bbb F_5$, simply $[\Bbb F_{125}:\Bbb F_{25}]$ would have to be $3/2$, which is absurd. – Marc van Leeuwen Nov 10 '12 at 09:13
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So if there is a subfield G of size 25, you can remove the zero element and it becomes a group under multiplication. Specifically, it is a subgroup of F-{0}, but F-{0}=124, and 24 doesn't divide 124. All subgroup sizes must divide the size of the group they come from, therefore you can never have a subfield of size 25.
Allan Henriques
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