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Prove $\mathbb{F}_{16}$ does not contain a subfield of order 8

Suppose $\mathbb{F}_{16}$ has a subfield $F$ of order 8. I want to use the tower law to arrive at a contradiction. That is, $4=|\mathbb{F}_{16}:\mathbb{Z}_2|=|\mathbb{F}_{16}:F||F:\mathbb{Z}_2|$. I don't know how to continue.

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    If $F$ is of order 8, what can you say about $[F:\mathbb{Z}_2]$? – 2017-02-02
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    Suppose $F$ has 8 elements. Any finite dimensional $F$-vector space (say, of dimension $n$) has $8^n$ elements. Now, $16 \neq 8^n$ for all $n$. – 2017-02-02

2 Answers 2

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We have $[F:\mathbb{F}_2]=[\mathbb{F}_{2^3}:\mathbb{F}_2]=3$, which does not divide $4$. Hence it is impossible.

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    Why $F$ is $\mathbb{F}_{2^3}$? – 2017-02-02
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    There is only one field of order $8$, so $F$ is ismorphic to $\mathbb{F}_8$. – 2017-02-02
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The multiplicative group of $\mathbb{F}_{16}$ has order $15$ while that of $\mathbb{F}_8$ has order $7$. Now you can use the Lagrange's theorem.