If $|G| = 10$ and $G$ is abelian, show that $G$ is cyclic

Question

What I have tried:

suppose $G$ is not cyclic, and let $\langle\,a\,\rangle$ be a cyclic subgroup of $G$ then $|a|\in\{1, 2, 5, 10\}$ by Lagrange's theorem. But I am stuck here and don't know how to proceed.

Any hints?

Answer 1

Probably the least involved proof. Let $L=\langle a \rangle $ be a nontrivial cyclic subgroup of $G$. Certainly $|L|>1$. If $|L|=10$ we are done. So by Lagrange $L$ is of order $2$ or $5$ so the order of $a$ is $2$ or $5$. Then the order of $G/L$ is $5$ or $2$ which is prime. Hence $G/L=\langle bL\rangle$ is cyclic of order $p=5$ or $2$. Then $b^p$ is in $L$. If $b^p\ne 1$ then $b$ has order $10$ because $b^p$ has order $10/p$ and we are done. If $b^p=1$ then $ab$ has order $10$ because $(ab)^p=a^p$ has order $10/p$.

Answer 2

By Cauchy's Theorem, $G$ has an element $a$ of order $2$ and an element $b$ of order $5$. Since $G$ is abelian, we have, in particular, that $ab=ba$. Indeed, $(ab)^{10}=a^{10}b^{10}=e$, and since $\gcd(2,5)=1$, the order of $ab$ is $10$.

Can you continue from here?

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Alternatively, since $10=2\times 5$, by a classification result, namely Theorem 7.3 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)", we have $G\cong \Bbb Z_{2\times 5}$ or $G\cong D_5$; but $G$ is abelian. Hence $G=\Bbb Z_{10}$ is cyclic.