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If we have a group $G$ s.t $|G|=15$. We also know $H \triangleleft G$ and that $H \cong \mathbb{Z}_3$.

Can we prove that $$ G \cong \mathbb{Z}_{15}? $$

Shaun
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Edge316
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  • Yes we can! Consider first $G/H$, it turns out to be isomorphic to $\mathbb{Z}_5$. And that means, there's a subgroup of $G$ of order $5$. But since $5$ is prime, this subgroup is cyclic. – Fakemistake Apr 04 '21 at 10:27

2 Answers2

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Since $|G| = 3 \cdot 5$ and $3 \not\mid (5-1)$ by classification theorem (Classify all groups with order $pq$, $p$ is not equal to $q$) you know that $G \simeq \mathbb{Z}_3 \times \mathbb{Z}_5 \simeq \mathbb{Z}_{15}$

jacopoburelli
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You can prove that the Sylow subgroups are unique, that is normal, when you have a group of order $pq, p\lt q,p\nmid q-1$. Then it follows that $G\cong\mathbb Z_p\times\mathbb Z_q\cong\mathbb Z_{pq}$.