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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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You'll never prove it, it’s not true. – Lubin Nov 10 '12 at 5:32
Consider objects of the form $x^5$, where $x$ ranges over the big field. – André Nicolas Nov 10 '12 at 5:34
up vote 8 down vote accepted

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?

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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 7:39
You 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 9:13

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