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Q : Let $p$ be a prime and let $F$ be a field. Let $K/F$ be a Galois extension whose Galois group is a $p$-group. Such extension is called a $p$-extension.\\

Let $L/K$ be a $p$-extension. Prove that the Galois closure of $L$ over $F$ is a $p$-extension of $F$.\\

A : Since $L/K$ and $K/F$ are $p$-extensions, $L/F$ is Galois. Since $L$ is separable, by the primitive element theorem, $L =  F(\alpha)$. Since $K/F$ is Galois, $m_\alpha\in F[x]$ splits over $K$ into a product of $n$ irreducibles of the same degree $d$. Since $dn = p^a$ for some integer $a$, $d$ is a power of $p$ so that $K(\beta)/K$ is a $p$-power degree for any root $\beta$ of $m_\alpha$. Define an isomorphism $\varphi:K(\alpha)\to K(\beta)$ by $\alpha\mapsto\beta$ (since they have the same degree in $K$). Then there is an induced isomorphism $Aut(K(\alpha)/K)\simeq Aut(K(\beta)/K)$ by $\sigma\mapsto \varphi\sigma\varphi^{-1}$. Here, $L/K = K(\alpha)/K$ is Galois by assumption, so $K(\beta)/K$ is Galois. Note that the Galois closure is the composite field of $K(\beta_i)$'s where $\beta_i$ for $1\leq i\leq l$ are the roots of $m_\alpha$ since it's a splitting field of $m_\alpha$ which is a separable polynomial. Now, $Gal(K(\beta_1)\cdots K(\beta_l)/K)<Gal(K(\beta_1)/K)\times\cdots\times Gal(K(\beta_l)/K)$ and each $Gal(K(\beta_i)/K)$ is a $p$-group, $Gal(K(\beta_1)\cdots K(\beta_l)/K)$ is a $p$-group. Since $K(\beta_1)\cdots K(\beta_l)$ is a splitting field of $m_\alpha$ over $F$, $K(\beta_1)\cdots K(\beta_l)/F$ is Galois whose degree is $p$ i.e. $p$-extension.