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Write $A=Aut(C_{7^3})$ as a direct product of cyclic groups of a prime order

$$A \cong C_{{p_1}^{m_1}} \times \ldots C_{{p_n}^{m_n}}$$

where p are prime numbers

there is a theorem that if $A=Aut(C_{p^e})$

then $A \cong S \times T$ where T is cyclic and $\lvert T \rvert = p-1 $ and $\lvert S \rvert = p^{e-1} $ and $S$ is generated by $\alpha_{p+1}$ where $\alpha_{p+1}(g) = g^{p+1}$

but Im not sure how to finish up from here

Daniel Katzan
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  • You probably want "A \cong S \ltimes T" ($A \cong S \ltimes T$). Otherwise, your description of $S$ being generated by an automorphism of some group containing $g$s is ... strange. – Eric Towers May 24 '16 at 08:28

1 Answers1

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We know that $Aut(C_n)$ is isomorphic to $U(n)$ of order $\phi(n)$, see here. For $n=7^3$ we have $\phi(n)=294=2\cdot 3\cdot 7^2$. Hence $A$ is a direct product of $C_2$, $C_3$ and $C_{49}$, because $U(n)$ is cyclic for prime powers $n=p^k$ with $p>2$, see here. So $A\cong C_{294}\cong C_2\times C_3\times C_{49}$.

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Dietrich Burde
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