In this exercise I had to find the Galois group of the splitting field $K $ of $(x^5-2)(x^2-5)$. I proved that this group is a semidirect product $\mathbb {Z}/5\mathbb {Z}/ \rtimes_\phi (\mathbb {Z}/5\mathbb {Z})^*,$ where $\mathbb {Z}/5\mathbb {Z}\cong Aut (K/\mathbb {Q}(\zeta_5))=<\tau>,$ and $(\mathbb {Z}/5\mathbb {Z})^*\cong Aut (K/\mathbb {Q}(\sqrt [5]{2}))=<\sigma>,$ with $\tau (\sqrt [5]{2})=\zeta_5\sqrt [5]{2}, \sigma (\zeta_5)=\zeta_5^2.$ Now, I'd like to describe $\phi:(\mathbb {Z}/5\mathbb {Z})^*\to Aut(\mathbb {Z}/5\mathbb {Z}).$ How can I do this?
A simple question on semidirect products
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abstract-algebra
1 Answers
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In an internal semidirect product $N \rtimes_\varphi H$, an element $h \in H$ (the non-normal factor) acts on $N$ by conjugation: $$ \varphi(h)(n) = h \cdot n = h n h^{-1} \, . $$ (See Dummit and Foote, Theorem $10$, $\S5.5$, p. $176$.) Thus, in your case, \begin{align*} \sigma \cdot \tau = \sigma \circ \tau \circ \sigma^{-1}: &\sqrt[5]{2} \overset{\sigma^{-1}}{\longmapsto} \sqrt[5]{2} \overset{\tau}{\longmapsto} \zeta_5 \sqrt[5]{2} \overset{\sigma}{\longmapsto} \zeta_5^2 \sqrt[5]{2}\\ & \zeta_5 \overset{\sigma^{-1}}{\longmapsto} \zeta_5^3 \overset{\tau}{\longmapsto} \zeta_5^3 \overset{\sigma}{\longmapsto} (\zeta_5^2)^3 = \zeta_5 \end{align*} so $\sigma \circ \tau \circ \sigma^{-1} = \tau^2$. Then $$ \sigma \circ \tau^k \circ \sigma^{-1} = (\sigma \circ \tau \circ \sigma^{-1})^k = (\tau^2)^k = \tau^{2k} $$ and in general $$ \sigma^j \circ \tau^k \circ \sigma^{-j} = \tau^{2^j k} \, . $$
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0Yes, but I'd like to see the elements as automorphisms of K.. how can I do it knowing this fact? – 2017-01-12
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0Does my edit not address this? I computed the action of $\sigma \circ \tau \circ \sigma^{-1}$ on the generators of $K$. You can do this similarly for $\sigma^{j} \circ \tau^{k} \circ \sigma^{-j}$. – 2017-01-12
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0Thanks a lot, is the elements you've indicated $\sigma^j\cdot\tau^k$? – 2017-01-12
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0I updated the answer to include $\sigma^j \cdot \tau^k$. – 2017-01-12